Termination w.r.t. Q proof of AProVE_07

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS

↳1 Overlay + Local Confluence (⇔, 0 ms)

↳2 QTRS

↳3 DependencyPairsProof (⇔, 0 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 AND

    ↳7 QDP

        ↳8 UsableRulesProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 QReductionProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 QDPSizeChangeProof (⇔, 0 ms)

        ↳13 YES

    ↳14 QDP

        ↳15 UsableRulesProof (⇔, 0 ms)

        ↳16 QDP

        ↳17 QReductionProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 TransformationProof (⇔, 0 ms)

        ↳20 QDP

        ↳21 TransformationProof (⇔, 0 ms)

        ↳22 QDP

        ↳23 DependencyGraphProof (⇔, 0 ms)

        ↳24 QDP

        ↳25 TransformationProof (⇔, 0 ms)

        ↳26 QDP

        ↳27 TransformationProof (⇔, 0 ms)

        ↳28 QDP

        ↳29 TransformationProof (⇔, 0 ms)

        ↳30 QDP

        ↳31 TransformationProof (⇔, 0 ms)

        ↳32 QDP

        ↳33 TransformationProof (⇔, 0 ms)

        ↳34 QDP

        ↳35 TransformationProof (⇔, 0 ms)

        ↳36 QDP

        ↳37 TransformationProof (⇔, 0 ms)

        ↳38 QDP

        ↳39 TransformationProof (⇔, 0 ms)

        ↳40 QDP

        ↳41 TransformationProof (⇔, 0 ms)

        ↳42 QDP

        ↳43 TransformationProof (⇔, 0 ms)

        ↳44 QDP

        ↳45 TransformationProof (⇔, 0 ms)

        ↳46 QDP

        ↳47 TransformationProof (⇔, 0 ms)

        ↳48 QDP

        ↳49 TransformationProof (⇔, 0 ms)

        ↳50 QDP

        ↳51 TransformationProof (⇔, 0 ms)

        ↳52 QDP

        ↳53 TransformationProof (⇔, 0 ms)

        ↳54 QDP

        ↳55 TransformationProof (⇔, 0 ms)

        ↳56 QDP

        ↳57 TransformationProof (⇔, 0 ms)

        ↳58 QDP

        ↳59 TransformationProof (⇔, 0 ms)

        ↳60 QDP

        ↳61 TransformationProof (⇔, 0 ms)

        ↳62 QDP

        ↳63 TransformationProof (⇔, 0 ms)

        ↳64 QDP

        ↳65 TransformationProof (⇔, 0 ms)

        ↳66 QDP

        ↳67 DependencyGraphProof (⇔, 0 ms)

        ↳68 QDP

        ↳69 TransformationProof (⇔, 0 ms)

        ↳70 QDP

        ↳71 TransformationProof (⇔, 0 ms)

        ↳72 QDP

        ↳73 TransformationProof (⇔, 0 ms)

        ↳74 QDP

        ↳75 TransformationProof (⇔, 0 ms)

        ↳76 QDP

        ↳77 TransformationProof (⇔, 0 ms)

        ↳78 QDP

        ↳79 UsableRulesProof (⇔, 0 ms)

        ↳80 QDP

        ↳81 TransformationProof (⇔, 0 ms)

        ↳82 QDP

        ↳83 UsableRulesProof (⇔, 0 ms)

        ↳84 QDP

        ↳85 QReductionProof (⇔, 0 ms)

        ↳86 QDP

        ↳87 TransformationProof (⇔, 0 ms)

        ↳88 QDP

        ↳89 TransformationProof (⇔, 0 ms)

        ↳90 QDP

        ↳91 TransformationProof (⇔, 0 ms)

        ↳92 QDP

        ↳93 TransformationProof (⇔, 0 ms)

        ↳94 QDP

        ↳95 TransformationProof (⇔, 0 ms)

        ↳96 QDP

        ↳97 TransformationProof (⇔, 0 ms)

        ↳98 QDP

        ↳99 TransformationProof (⇔, 0 ms)

        ↳100 QDP

        ↳101 TransformationProof (⇔, 0 ms)

        ↳102 QDP

        ↳103 TransformationProof (⇔, 0 ms)

        ↳104 QDP

        ↳105 TransformationProof (⇔, 0 ms)

        ↳106 QDP

        ↳107 TransformationProof (⇔, 0 ms)

        ↳108 QDP

        ↳109 TransformationProof (⇔, 0 ms)

        ↳110 QDP

        ↳111 TransformationProof (⇔, 0 ms)

        ↳112 QDP

        ↳113 DependencyGraphProof (⇔, 0 ms)

        ↳114 AND

            ↳115 QDP

                ↳116 TransformationProof (⇔, 0 ms)

                ↳117 QDP

                ↳118 TransformationProof (⇔, 0 ms)

                ↳119 QDP

                ↳120 TransformationProof (⇔, 0 ms)

                ↳121 QDP

                ↳122 TransformationProof (⇔, 0 ms)

                ↳123 QDP

                ↳124 DependencyGraphProof (⇔, 0 ms)

                ↳125 QDP

                ↳126 TransformationProof (⇔, 0 ms)

                ↳127 QDP

                ↳128 TransformationProof (⇔, 0 ms)

                ↳129 QDP

                ↳130 TransformationProof (⇔, 0 ms)

                ↳131 QDP

                ↳132 TransformationProof (⇔, 0 ms)

                ↳133 QDP

                ↳134 UsableRulesProof (⇔, 0 ms)

                ↳135 QDP

                ↳136 TransformationProof (⇔, 0 ms)

                ↳137 QDP

                ↳138 TransformationProof (⇔, 0 ms)

                ↳139 QDP

                ↳140 QDPQMonotonicMRRProof (⇔, 4 ms)

                ↳141 QDP

                ↳142 DependencyGraphProof (⇔, 0 ms)

                ↳143 TRUE

            ↳144 QDP

                ↳145 UsableRulesProof (⇔, 0 ms)

                ↳146 QDP

                ↳147 QDPSizeChangeProof (⇔, 0 ms)

                ↳148 YES

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
average(x, y) → if(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
if(true, b1, b2, b3, x, y) → if2(b1, b2, b3, x, y)
if(false, b1, b2, b3, x, y) → average(p(x), s(y))
if2(true, b2, b3, x, y) → 0
if2(false, b2, b3, x, y) → if3(b2, b3, x, y)
if3(true, b3, x, y) → 0
if3(false, b3, x, y) → if4(b3, x, y)
if4(true, x, y) → s(0)
if4(false, x, y) → average(s(x), p(p(y)))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
average(x, y) → if(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
if(true, b1, b2, b3, x, y) → if2(b1, b2, b3, x, y)
if(false, b1, b2, b3, x, y) → average(p(x), s(y))
if2(true, b2, b3, x, y) → 0
if2(false, b2, b3, x, y) → if3(b2, b3, x, y)
if3(true, b3, x, y) → 0
if3(false, b3, x, y) → if4(b3, x, y)
if4(true, x, y) → s(0)
if4(false, x, y) → average(s(x), p(p(y)))

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
average(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, x3)
if2(false, x0, x1, x2, x3)
if3(true, x0, x1, x2)
if3(false, x0, x1, x2)
if4(true, x0, x1)
if4(false, x0, x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s(x), s(y)) → LE(x, y)
AVERAGE(x, y) → IF(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
AVERAGE(x, y) → LE(x, 0)
AVERAGE(x, y) → LE(y, 0)
AVERAGE(x, y) → LE(y, s(0))
AVERAGE(x, y) → LE(y, s(s(0)))
IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)
IF(false, b1, b2, b3, x, y) → AVERAGE(p(x), s(y))
IF(false, b1, b2, b3, x, y) → P(x)
IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
IF3(false, b3, x, y) → IF4(b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
IF4(false, x, y) → P(p(y))
IF4(false, x, y) → P(y)

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
average(x, y) → if(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
if(true, b1, b2, b3, x, y) → if2(b1, b2, b3, x, y)
if(false, b1, b2, b3, x, y) → average(p(x), s(y))
if2(true, b2, b3, x, y) → 0
if2(false, b2, b3, x, y) → if3(b2, b3, x, y)
if3(true, b3, x, y) → 0
if3(false, b3, x, y) → if4(b3, x, y)
if4(true, x, y) → s(0)
if4(false, x, y) → average(s(x), p(p(y)))

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
average(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, x3)
if2(false, x0, x1, x2, x3)
if3(true, x0, x1, x2)
if3(false, x0, x1, x2)
if4(true, x0, x1)
if4(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 7 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
average(x, y) → if(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
if(true, b1, b2, b3, x, y) → if2(b1, b2, b3, x, y)
if(false, b1, b2, b3, x, y) → average(p(x), s(y))
if2(true, b2, b3, x, y) → 0
if2(false, b2, b3, x, y) → if3(b2, b3, x, y)
if3(true, b3, x, y) → 0
if3(false, b3, x, y) → if4(b3, x, y)
if4(true, x, y) → s(0)
if4(false, x, y) → average(s(x), p(p(y)))

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
average(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, x3)
if2(false, x0, x1, x2, x3)
if3(true, x0, x1, x2)
if3(false, x0, x1, x2)
if4(true, x0, x1)
if4(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s(x), s(y)) → LE(x, y)

R is empty.
The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
average(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, x3)
if2(false, x0, x1, x2, x3)
if3(true, x0, x1, x2)
if3(false, x0, x1, x2)
if4(true, x0, x1)
if4(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
average(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, x3)
if2(false, x0, x1, x2, x3)
if3(true, x0, x1, x2)
if3(false, x0, x1, x2)
if4(true, x0, x1)
if4(false, x0, x1)

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LE(s(x), s(y)) → LE(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

LE(s(x), s(y)) → LE(x, y)
The graph contains the following edges 1 > 1, 2 > 2

(13) YES

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(x, y) → IF(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)
IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
IF3(false, b3, x, y) → IF4(b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
IF(false, b1, b2, b3, x, y) → AVERAGE(p(x), s(y))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
average(x, y) → if(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
if(true, b1, b2, b3, x, y) → if2(b1, b2, b3, x, y)
if(false, b1, b2, b3, x, y) → average(p(x), s(y))
if2(true, b2, b3, x, y) → 0
if2(false, b2, b3, x, y) → if3(b2, b3, x, y)
if3(true, b3, x, y) → 0
if3(false, b3, x, y) → if4(b3, x, y)
if4(true, x, y) → s(0)
if4(false, x, y) → average(s(x), p(p(y)))

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
average(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, x3)
if2(false, x0, x1, x2, x3)
if3(true, x0, x1, x2)
if3(false, x0, x1, x2)
if4(true, x0, x1)
if4(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(x, y) → IF(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)
IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
IF3(false, b3, x, y) → IF4(b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
IF(false, b1, b2, b3, x, y) → AVERAGE(p(x), s(y))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
average(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, x3)
if2(false, x0, x1, x2, x3)
if3(true, x0, x1, x2)
if3(false, x0, x1, x2)
if4(true, x0, x1)
if4(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

average(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, x3)
if2(false, x0, x1, x2, x3)
if3(true, x0, x1, x2)
if3(false, x0, x1, x2)
if4(true, x0, x1)
if4(false, x0, x1)

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(x, y) → IF(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)
IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
IF3(false, b3, x, y) → IF4(b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
IF(false, b1, b2, b3, x, y) → AVERAGE(p(x), s(y))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(19) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule AVERAGE(x, y) → IF(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y) at position [0] we obtained the following new rules [LPAR04]:

AVERAGE(0, y1) → IF(true, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), 0, y1) → AVERAGE(0, y1) → IF(true, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), 0, y1)
AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1) → AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1)

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)
IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
IF3(false, b3, x, y) → IF4(b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
IF(false, b1, b2, b3, x, y) → AVERAGE(p(x), s(y))
AVERAGE(0, y1) → IF(true, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), 0, y1)
AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1)

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(21) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF(false, b1, b2, b3, x, y) → AVERAGE(p(x), s(y)) at position [0] we obtained the following new rules [LPAR04]:

IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4)) → IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
IF(false, y0, y1, y2, 0, y4) → AVERAGE(0, s(y4)) → IF(false, y0, y1, y2, 0, y4) → AVERAGE(0, s(y4))

(22) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)
IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
IF3(false, b3, x, y) → IF4(b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
AVERAGE(0, y1) → IF(true, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), 0, y1)
AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1)
IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
IF(false, y0, y1, y2, 0, y4) → AVERAGE(0, s(y4))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(24) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
IF3(false, b3, x, y) → IF4(b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1)
IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
AVERAGE(0, y1) → IF(true, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), 0, y1)
IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(25) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule AVERAGE(0, y1) → IF(true, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), 0, y1) we obtained the following new rules [LPAR04]:

AVERAGE(0, s(z4)) → IF(true, le(s(z4), 0), le(s(z4), s(0)), le(s(z4), s(s(0))), 0, s(z4)) → AVERAGE(0, s(z4)) → IF(true, le(s(z4), 0), le(s(z4), s(0)), le(s(z4), s(s(0))), 0, s(z4))

(26) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
IF3(false, b3, x, y) → IF4(b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1)
IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)
AVERAGE(0, s(z4)) → IF(true, le(s(z4), 0), le(s(z4), s(0)), le(s(z4), s(s(0))), 0, s(z4))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(27) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(0, s(z4)) → IF(true, le(s(z4), 0), le(s(z4), s(0)), le(s(z4), s(s(0))), 0, s(z4)) at position [1] we obtained the following new rules [LPAR04]:

AVERAGE(0, s(z4)) → IF(true, false, le(s(z4), s(0)), le(s(z4), s(s(0))), 0, s(z4)) → AVERAGE(0, s(z4)) → IF(true, false, le(s(z4), s(0)), le(s(z4), s(s(0))), 0, s(z4))

(28) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
IF3(false, b3, x, y) → IF4(b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1)
IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)
AVERAGE(0, s(z4)) → IF(true, false, le(s(z4), s(0)), le(s(z4), s(s(0))), 0, s(z4))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(29) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(0, s(z4)) → IF(true, false, le(s(z4), s(0)), le(s(z4), s(s(0))), 0, s(z4)) at position [2] we obtained the following new rules [LPAR04]:

AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(s(z4), s(s(0))), 0, s(z4)) → AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(s(z4), s(s(0))), 0, s(z4))

(30) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
IF3(false, b3, x, y) → IF4(b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1)
IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)
AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(s(z4), s(s(0))), 0, s(z4))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(31) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(s(z4), s(s(0))), 0, s(z4)) at position [3] we obtained the following new rules [LPAR04]:

AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4)) → AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))

(32) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
IF3(false, b3, x, y) → IF4(b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1)
IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)
AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(33) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y) we obtained the following new rules [LPAR04]:

IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0)) → IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))

(34) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
IF3(false, b3, x, y) → IF4(b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1)
IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(35) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y) we obtained the following new rules [LPAR04]:

IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2)) → IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))

(36) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF3(false, b3, x, y) → IF4(b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1)
IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(37) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF3(false, b3, x, y) → IF4(b3, x, y) we obtained the following new rules [LPAR04]:

IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2)) → IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))

(38) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1)
IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(39) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF4(false, x, y) → AVERAGE(s(x), p(p(y))) we obtained the following new rules [LPAR04]:

IF4(false, 0, s(z1)) → AVERAGE(s(0), p(p(s(z1)))) → IF4(false, 0, s(z1)) → AVERAGE(s(0), p(p(s(z1))))

(40) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1)
IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(p(s(z1))))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(41) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF4(false, 0, s(z1)) → AVERAGE(s(0), p(p(s(z1)))) at position [1,0] we obtained the following new rules [LPAR04]:

IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1)) → IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))

(42) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1)
IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(43) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule AVERAGE(s(x0), y1) → IF(false, le(y1, 0), le(y1, s(0)), le(y1, s(s(0))), s(x0), y1) we obtained the following new rules [LPAR04]:

AVERAGE(s(x0), s(z4)) → IF(false, le(s(z4), 0), le(s(z4), s(0)), le(s(z4), s(s(0))), s(x0), s(z4)) → AVERAGE(s(x0), s(z4)) → IF(false, le(s(z4), 0), le(s(z4), s(0)), le(s(z4), s(s(0))), s(x0), s(z4))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0) → AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)

(44) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(x0), s(z4)) → IF(false, le(s(z4), 0), le(s(z4), s(0)), le(s(z4), s(s(0))), s(x0), s(z4))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(45) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(x0), s(z4)) → IF(false, le(s(z4), 0), le(s(z4), s(0)), le(s(z4), s(s(0))), s(x0), s(z4)) at position [1] we obtained the following new rules [LPAR04]:

AVERAGE(s(x0), s(z4)) → IF(false, false, le(s(z4), s(0)), le(s(z4), s(s(0))), s(x0), s(z4)) → AVERAGE(s(x0), s(z4)) → IF(false, false, le(s(z4), s(0)), le(s(z4), s(s(0))), s(x0), s(z4))

(46) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
AVERAGE(s(x0), s(z4)) → IF(false, false, le(s(z4), s(0)), le(s(z4), s(s(0))), s(x0), s(z4))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(47) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(x0), s(z4)) → IF(false, false, le(s(z4), s(0)), le(s(z4), s(s(0))), s(x0), s(z4)) at position [2] we obtained the following new rules [LPAR04]:

AVERAGE(s(x0), s(z4)) → IF(false, false, le(z4, 0), le(s(z4), s(s(0))), s(x0), s(z4)) → AVERAGE(s(x0), s(z4)) → IF(false, false, le(z4, 0), le(s(z4), s(s(0))), s(x0), s(z4))

(48) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
AVERAGE(s(x0), s(z4)) → IF(false, false, le(z4, 0), le(s(z4), s(s(0))), s(x0), s(z4))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(49) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(x0), s(z4)) → IF(false, false, le(z4, 0), le(s(z4), s(s(0))), s(x0), s(z4)) at position [3] we obtained the following new rules [LPAR04]:

AVERAGE(s(x0), s(z4)) → IF(false, false, le(z4, 0), le(z4, s(0)), s(x0), s(z4)) → AVERAGE(s(x0), s(z4)) → IF(false, false, le(z4, 0), le(z4, s(0)), s(x0), s(z4))

(50) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4))
AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
AVERAGE(s(x0), s(z4)) → IF(false, false, le(z4, 0), le(z4, s(0)), s(x0), s(z4))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(51) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, y0, y1, y2, s(x0), y4) → AVERAGE(x0, s(y4)) we obtained the following new rules [LPAR04]:

IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0)) → IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF(false, false, y_0, y_1, s(z0), s(z1)) → AVERAGE(z0, s(s(z1))) → IF(false, false, y_0, y_1, s(z0), s(z1)) → AVERAGE(z0, s(s(z1)))

(52) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
AVERAGE(s(x0), s(z4)) → IF(false, false, le(z4, 0), le(z4, s(0)), s(x0), s(z4))
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF(false, false, y_0, y_1, s(z0), s(z1)) → AVERAGE(z0, s(s(z1)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(53) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule AVERAGE(s(x0), s(z4)) → IF(false, false, le(z4, 0), le(z4, s(0)), s(x0), s(z4)) we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1)) → AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, le(s(z3), 0), le(s(z3), s(0)), s(x0), s(s(z3))) → AVERAGE(s(x0), s(s(z3))) → IF(false, false, le(s(z3), 0), le(s(z3), s(0)), s(x0), s(s(z3)))

(54) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF(false, false, y_0, y_1, s(z0), s(z1)) → AVERAGE(z0, s(s(z1)))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, le(s(z3), 0), le(s(z3), s(0)), s(x0), s(s(z3)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(55) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(x0), s(s(z3))) → IF(false, false, le(s(z3), 0), le(s(z3), s(0)), s(x0), s(s(z3))) at position [2] we obtained the following new rules [LPAR04]:

AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(s(z3), s(0)), s(x0), s(s(z3))) → AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(s(z3), s(0)), s(x0), s(s(z3)))

(56) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF(false, false, y_0, y_1, s(z0), s(z1)) → AVERAGE(z0, s(s(z1)))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(s(z3), s(0)), s(x0), s(s(z3)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(57) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(s(z3), s(0)), s(x0), s(s(z3))) at position [3] we obtained the following new rules [LPAR04]:

AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3))) → AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))

(58) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF(false, false, y_0, y_1, s(z0), s(z1)) → AVERAGE(z0, s(s(z1)))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(59) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, false, y_0, y_1, s(z0), s(z1)) → AVERAGE(z0, s(s(z1))) we obtained the following new rules [LPAR04]:

IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3))) → IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1)))) → IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))

(60) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(61) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF2(false, z0, z1, 0, s(z2)) → IF3(z0, z1, 0, s(z2)) we obtained the following new rules [LPAR04]:

IF2(false, false, x1, 0, s(x2)) → IF3(false, x1, 0, s(x2)) → IF2(false, false, x1, 0, s(x2)) → IF3(false, x1, 0, s(x2))

(62) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
IF2(false, false, x1, 0, s(x2)) → IF3(false, x1, 0, s(x2))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(63) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF(true, false, y_0, y_1, 0, s(z0)) → IF2(false, y_0, y_1, 0, s(z0)) we obtained the following new rules [LPAR04]:

IF(true, false, false, x1, 0, s(x2)) → IF2(false, false, x1, 0, s(x2)) → IF(true, false, false, x1, 0, s(x2)) → IF2(false, false, x1, 0, s(x2))

(64) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
IF2(false, false, x1, 0, s(x2)) → IF3(false, x1, 0, s(x2))
IF(true, false, false, x1, 0, s(x2)) → IF2(false, false, x1, 0, s(x2))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(65) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule AVERAGE(0, s(z4)) → IF(true, false, le(z4, 0), le(z4, s(0)), 0, s(z4)) at position [2] we obtained the following new rules [LPAR04]:

AVERAGE(0, s(0)) → IF(true, false, true, le(0, s(0)), 0, s(0)) → AVERAGE(0, s(0)) → IF(true, false, true, le(0, s(0)), 0, s(0))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(s(x0), s(0)), 0, s(s(x0))) → AVERAGE(0, s(s(x0))) → IF(true, false, false, le(s(x0), s(0)), 0, s(s(x0)))

(66) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
IF2(false, false, x1, 0, s(x2)) → IF3(false, x1, 0, s(x2))
IF(true, false, false, x1, 0, s(x2)) → IF2(false, false, x1, 0, s(x2))
AVERAGE(0, s(0)) → IF(true, false, true, le(0, s(0)), 0, s(0))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(s(x0), s(0)), 0, s(s(x0)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(67) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(68) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(s(x0), s(0)), 0, s(s(x0)))
IF(true, false, false, x1, 0, s(x2)) → IF2(false, false, x1, 0, s(x2))
IF2(false, false, x1, 0, s(x2)) → IF3(false, x1, 0, s(x2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(69) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(0, s(s(x0))) → IF(true, false, false, le(s(x0), s(0)), 0, s(s(x0))) at position [3] we obtained the following new rules [LPAR04]:

AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0))) → AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))

(70) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF(true, false, false, x1, 0, s(x2)) → IF2(false, false, x1, 0, s(x2))
IF2(false, false, x1, 0, s(x2)) → IF3(false, x1, 0, s(x2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(71) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(true, false, false, x1, 0, s(x2)) → IF2(false, false, x1, 0, s(x2)) we obtained the following new rules [LPAR04]:

IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0))) → IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))

(72) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF2(false, false, x1, 0, s(x2)) → IF3(false, x1, 0, s(x2))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(73) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF2(false, false, x1, 0, s(x2)) → IF3(false, x1, 0, s(x2)) we obtained the following new rules [LPAR04]:

IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1))) → IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))

(74) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(75) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF3(false, z1, 0, s(z2)) → IF4(z1, 0, s(z2)) we obtained the following new rules [LPAR04]:

IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))

(76) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1))
AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(77) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF4(false, 0, s(z1)) → AVERAGE(s(0), p(z1)) we obtained the following new rules [LPAR04]:

IF4(false, 0, s(s(z1))) → AVERAGE(s(0), p(s(z1))) → IF4(false, 0, s(s(z1))) → AVERAGE(s(0), p(s(z1)))

(78) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), p(s(z1)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(79) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(80) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), p(s(z1)))

The TRS R consists of the following rules:

p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(81) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF4(false, 0, s(s(z1))) → AVERAGE(s(0), p(s(z1))) at position [1] we obtained the following new rules [LPAR04]:

IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1) → IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)

(82) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)

The TRS R consists of the following rules:

p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(83) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(84) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(s(x0))
p(0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(85) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

p(s(x0))
p(0)

(86) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(87) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF(false, y_0, y_1, y_2, s(0), z0) → AVERAGE(0, s(z0)) we obtained the following new rules [LPAR04]:

IF(false, x0, x1, x2, s(0), s(y_0)) → AVERAGE(0, s(s(y_0))) → IF(false, x0, x1, x2, s(0), s(y_0)) → AVERAGE(0, s(s(y_0)))

(88) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0)
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
IF(false, x0, x1, x2, s(0), s(y_0)) → AVERAGE(0, s(s(y_0)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(89) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule AVERAGE(s(0), y_0) → IF(false, le(y_0, 0), le(y_0, s(0)), le(y_0, s(s(0))), s(0), y_0) we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(y_4)) → IF(false, le(s(y_4), 0), le(s(y_4), s(0)), le(s(y_4), s(s(0))), s(0), s(y_4)) → AVERAGE(s(0), s(y_4)) → IF(false, le(s(y_4), 0), le(s(y_4), s(0)), le(s(y_4), s(s(0))), s(0), s(y_4))
AVERAGE(s(0), s(s(y_5))) → IF(false, le(s(s(y_5)), 0), le(s(s(y_5)), s(0)), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5))) → AVERAGE(s(0), s(s(y_5))) → IF(false, le(s(s(y_5)), 0), le(s(s(y_5)), s(0)), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5)))

(90) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
IF(false, x0, x1, x2, s(0), s(y_0)) → AVERAGE(0, s(s(y_0)))
AVERAGE(s(0), s(y_4)) → IF(false, le(s(y_4), 0), le(s(y_4), s(0)), le(s(y_4), s(s(0))), s(0), s(y_4))
AVERAGE(s(0), s(s(y_5))) → IF(false, le(s(s(y_5)), 0), le(s(s(y_5)), s(0)), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(91) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(0), s(y_4)) → IF(false, le(s(y_4), 0), le(s(y_4), s(0)), le(s(y_4), s(s(0))), s(0), s(y_4)) at position [1] we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(y_4)) → IF(false, false, le(s(y_4), s(0)), le(s(y_4), s(s(0))), s(0), s(y_4)) → AVERAGE(s(0), s(y_4)) → IF(false, false, le(s(y_4), s(0)), le(s(y_4), s(s(0))), s(0), s(y_4))

(92) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
IF(false, x0, x1, x2, s(0), s(y_0)) → AVERAGE(0, s(s(y_0)))
AVERAGE(s(0), s(s(y_5))) → IF(false, le(s(s(y_5)), 0), le(s(s(y_5)), s(0)), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5)))
AVERAGE(s(0), s(y_4)) → IF(false, false, le(s(y_4), s(0)), le(s(y_4), s(s(0))), s(0), s(y_4))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(93) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(0), s(s(y_5))) → IF(false, le(s(s(y_5)), 0), le(s(s(y_5)), s(0)), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5))) at position [1] we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(s(y_5))) → IF(false, false, le(s(s(y_5)), s(0)), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5))) → AVERAGE(s(0), s(s(y_5))) → IF(false, false, le(s(s(y_5)), s(0)), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5)))

(94) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
IF(false, x0, x1, x2, s(0), s(y_0)) → AVERAGE(0, s(s(y_0)))
AVERAGE(s(0), s(y_4)) → IF(false, false, le(s(y_4), s(0)), le(s(y_4), s(s(0))), s(0), s(y_4))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, le(s(s(y_5)), s(0)), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(95) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(0), s(y_4)) → IF(false, false, le(s(y_4), s(0)), le(s(y_4), s(s(0))), s(0), s(y_4)) at position [2] we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(y_4)) → IF(false, false, le(y_4, 0), le(s(y_4), s(s(0))), s(0), s(y_4)) → AVERAGE(s(0), s(y_4)) → IF(false, false, le(y_4, 0), le(s(y_4), s(s(0))), s(0), s(y_4))

(96) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
IF(false, x0, x1, x2, s(0), s(y_0)) → AVERAGE(0, s(s(y_0)))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, le(s(s(y_5)), s(0)), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5)))
AVERAGE(s(0), s(y_4)) → IF(false, false, le(y_4, 0), le(s(y_4), s(s(0))), s(0), s(y_4))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(97) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(0), s(s(y_5))) → IF(false, false, le(s(s(y_5)), s(0)), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5))) at position [2] we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(s(y_5))) → IF(false, false, le(s(y_5), 0), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5))) → AVERAGE(s(0), s(s(y_5))) → IF(false, false, le(s(y_5), 0), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5)))

(98) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
IF(false, x0, x1, x2, s(0), s(y_0)) → AVERAGE(0, s(s(y_0)))
AVERAGE(s(0), s(y_4)) → IF(false, false, le(y_4, 0), le(s(y_4), s(s(0))), s(0), s(y_4))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, le(s(y_5), 0), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(99) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(0), s(y_4)) → IF(false, false, le(y_4, 0), le(s(y_4), s(s(0))), s(0), s(y_4)) at position [3] we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(y_4)) → IF(false, false, le(y_4, 0), le(y_4, s(0)), s(0), s(y_4)) → AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))

(100) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
IF(false, x0, x1, x2, s(0), s(y_0)) → AVERAGE(0, s(s(y_0)))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, le(s(y_5), 0), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(101) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(0), s(s(y_5))) → IF(false, false, le(s(y_5), 0), le(s(s(y_5)), s(s(0))), s(0), s(s(y_5))) at position [2] we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(s(s(y_5)), s(s(0))), s(0), s(s(y_5))) → AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(s(s(y_5)), s(s(0))), s(0), s(s(y_5)))

(102) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
IF(false, x0, x1, x2, s(0), s(y_0)) → AVERAGE(0, s(s(y_0)))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(s(s(y_5)), s(s(0))), s(0), s(s(y_5)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(103) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(s(s(y_5)), s(s(0))), s(0), s(s(y_5))) at position [3] we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(s(y_5), s(0)), s(0), s(s(y_5))) → AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(s(y_5), s(0)), s(0), s(s(y_5)))

(104) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
IF(false, x0, x1, x2, s(0), s(y_0)) → AVERAGE(0, s(s(y_0)))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(s(y_5), s(0)), s(0), s(s(y_5)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(105) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(s(y_5), s(0)), s(0), s(s(y_5))) at position [3] we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5))) → AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))

(106) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
IF(false, x0, x1, x2, s(0), s(y_0)) → AVERAGE(0, s(s(y_0)))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(107) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, x0, x1, x2, s(0), s(y_0)) → AVERAGE(0, s(s(y_0))) we obtained the following new rules [LPAR04]:

IF(false, false, y_0, y_1, s(0), s(z0)) → AVERAGE(0, s(s(z0))) → IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1)))) → IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))

(108) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1))))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(109) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF(false, false, false, y_0, s(z0), s(s(z1))) → AVERAGE(z0, s(s(s(z1)))) we obtained the following new rules [LPAR04]:

IF(false, false, false, x0, s(s(0)), s(s(x2))) → AVERAGE(s(0), s(s(s(x2)))) → IF(false, false, false, x0, s(s(0)), s(s(x2))) → AVERAGE(s(0), s(s(s(x2))))
IF(false, false, false, x0, s(s(y_0)), s(s(x2))) → AVERAGE(s(y_0), s(s(s(x2)))) → IF(false, false, false, x0, s(s(y_0)), s(s(x2))) → AVERAGE(s(y_0), s(s(s(x2))))
IF(false, false, false, x0, s(0), s(s(x2))) → AVERAGE(0, s(s(s(x2)))) → IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))

(110) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3)))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))
IF(false, false, false, x0, s(s(0)), s(s(x2))) → AVERAGE(s(0), s(s(s(x2))))
IF(false, false, false, x0, s(s(y_0)), s(s(x2))) → AVERAGE(s(y_0), s(s(s(x2))))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(111) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule AVERAGE(s(x0), s(s(z3))) → IF(false, false, false, le(z3, 0), s(x0), s(s(z3))) we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(s(x1))) → IF(false, false, false, le(x1, 0), s(0), s(s(x1))) → AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
AVERAGE(s(s(0)), s(s(x1))) → IF(false, false, false, le(x1, 0), s(s(0)), s(s(x1))) → AVERAGE(s(s(0)), s(s(x1))) → IF(false, false, false, le(x1, 0), s(s(0)), s(s(x1)))
AVERAGE(s(s(y_4)), s(s(x1))) → IF(false, false, false, le(x1, 0), s(s(y_4)), s(s(x1))) → AVERAGE(s(s(y_4)), s(s(x1))) → IF(false, false, false, le(x1, 0), s(s(y_4)), s(s(x1)))

(112) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))
IF(false, false, false, x0, s(s(0)), s(s(x2))) → AVERAGE(s(0), s(s(s(x2))))
IF(false, false, false, x0, s(s(y_0)), s(s(x2))) → AVERAGE(s(y_0), s(s(s(x2))))
AVERAGE(s(s(0)), s(s(x1))) → IF(false, false, false, le(x1, 0), s(s(0)), s(s(x1)))
AVERAGE(s(s(y_4)), s(s(x1))) → IF(false, false, false, le(x1, 0), s(s(y_4)), s(s(x1)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(113) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(114) Complex Obligation (AND)

(115) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(116) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF3(false, z0, 0, s(s(z1))) → IF4(z0, 0, s(s(z1))) we obtained the following new rules [LPAR04]:

IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))

(117) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(118) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF2(false, false, z0, 0, s(s(z1))) → IF3(false, z0, 0, s(s(z1))) we obtained the following new rules [LPAR04]:

IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1)))

(119) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))
IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(120) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF(true, false, false, y_0, 0, s(s(z0))) → IF2(false, false, y_0, 0, s(s(z0))) we obtained the following new rules [LPAR04]:

IF(true, false, false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1))) → IF(true, false, false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1)))

(121) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))
IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1)))
IF(true, false, false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(122) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule AVERAGE(0, s(s(x0))) → IF(true, false, false, le(x0, 0), 0, s(s(x0))) at position [3] we obtained the following new rules [LPAR04]:

AVERAGE(0, s(s(0))) → IF(true, false, false, true, 0, s(s(0))) → AVERAGE(0, s(s(0))) → IF(true, false, false, true, 0, s(s(0)))
AVERAGE(0, s(s(s(x0)))) → IF(true, false, false, false, 0, s(s(s(x0)))) → AVERAGE(0, s(s(s(x0)))) → IF(true, false, false, false, 0, s(s(s(x0))))

(123) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))
IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1)))
IF(true, false, false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1)))
AVERAGE(0, s(s(0))) → IF(true, false, false, true, 0, s(s(0)))
AVERAGE(0, s(s(s(x0)))) → IF(true, false, false, false, 0, s(s(s(x0))))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(124) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(125) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3)))
AVERAGE(0, s(s(s(x0)))) → IF(true, false, false, false, 0, s(s(s(x0))))
IF(true, false, false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1)))
IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1)))
IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(126) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF(false, false, y_1, y_2, s(0), s(x3)) → AVERAGE(0, s(s(x3))) we obtained the following new rules [LPAR04]:

IF(false, false, x0, x1, s(0), s(s(y_0))) → AVERAGE(0, s(s(s(y_0)))) → IF(false, false, x0, x1, s(0), s(s(y_0))) → AVERAGE(0, s(s(s(y_0))))

(127) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1))
AVERAGE(0, s(s(s(x0)))) → IF(true, false, false, false, 0, s(s(s(x0))))
IF(true, false, false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1)))
IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1)))
IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))
IF(false, false, x0, x1, s(0), s(s(y_0))) → AVERAGE(0, s(s(s(y_0))))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(128) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule AVERAGE(s(0), s(x1)) → IF(false, false, le(x1, 0), le(x1, s(0)), s(0), s(x1)) we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(s(y_4))) → IF(false, false, le(s(y_4), 0), le(s(y_4), s(0)), s(0), s(s(y_4))) → AVERAGE(s(0), s(s(y_4))) → IF(false, false, le(s(y_4), 0), le(s(y_4), s(0)), s(0), s(s(y_4)))

(129) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(s(s(x0)))) → IF(true, false, false, false, 0, s(s(s(x0))))
IF(true, false, false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1)))
IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1)))
IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))
IF(false, false, x0, x1, s(0), s(s(y_0))) → AVERAGE(0, s(s(s(y_0))))
AVERAGE(s(0), s(s(y_4))) → IF(false, false, le(s(y_4), 0), le(s(y_4), s(0)), s(0), s(s(y_4)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(130) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(0), s(s(y_4))) → IF(false, false, le(s(y_4), 0), le(s(y_4), s(0)), s(0), s(s(y_4))) at position [2] we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(s(y_4))) → IF(false, false, false, le(s(y_4), s(0)), s(0), s(s(y_4))) → AVERAGE(s(0), s(s(y_4))) → IF(false, false, false, le(s(y_4), s(0)), s(0), s(s(y_4)))

(131) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(s(s(x0)))) → IF(true, false, false, false, 0, s(s(s(x0))))
IF(true, false, false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1)))
IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1)))
IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))
IF(false, false, x0, x1, s(0), s(s(y_0))) → AVERAGE(0, s(s(s(y_0))))
AVERAGE(s(0), s(s(y_4))) → IF(false, false, false, le(s(y_4), s(0)), s(0), s(s(y_4)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(132) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule AVERAGE(s(0), s(s(y_4))) → IF(false, false, false, le(s(y_4), s(0)), s(0), s(s(y_4))) at position [3] we obtained the following new rules [LPAR04]:

AVERAGE(s(0), s(s(y_4))) → IF(false, false, false, le(y_4, 0), s(0), s(s(y_4))) → AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))

(133) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(s(s(x0)))) → IF(true, false, false, false, 0, s(s(s(x0))))
IF(true, false, false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1)))
IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1)))
IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))
IF(false, false, x0, x1, s(0), s(s(y_0))) → AVERAGE(0, s(s(s(y_0))))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(134) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(135) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(s(s(x0)))) → IF(true, false, false, false, 0, s(s(s(x0))))
IF(true, false, false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1)))
IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1)))
IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))
IF(false, false, x0, x1, s(0), s(s(y_0))) → AVERAGE(0, s(s(s(y_0))))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(136) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, false, x0, x1, s(0), s(s(y_0))) → AVERAGE(0, s(s(s(y_0)))) we obtained the following new rules [LPAR04]:

IF(false, false, false, y_0, s(0), s(s(z0))) → AVERAGE(0, s(s(s(z0)))) → IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))

(137) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(s(s(x0)))) → IF(true, false, false, false, 0, s(s(s(x0))))
IF(true, false, false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1)))
IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1)))
IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))
IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1)
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(138) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF4(false, 0, s(s(z1))) → AVERAGE(s(0), z1) we obtained the following new rules [LPAR04]:

IF4(false, 0, s(s(s(s(y_0))))) → AVERAGE(s(0), s(s(y_0))) → IF4(false, 0, s(s(s(s(y_0))))) → AVERAGE(s(0), s(s(y_0)))

(139) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(s(s(x0)))) → IF(true, false, false, false, 0, s(s(s(x0))))
IF(true, false, false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1)))
IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1)))
IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))
IF4(false, 0, s(s(s(s(y_0))))) → AVERAGE(s(0), s(s(y_0)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(140) QDPQMonotonicMRRProof (EQUIVALENT transformation)

By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain.
Strictly oriented dependency pairs:

IF(true, false, false, false, 0, s(s(x1))) → IF2(false, false, false, 0, s(s(x1)))
IF4(false, 0, s(s(s(s(y_0))))) → AVERAGE(s(0), s(s(y_0)))

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1
POL(AVERAGE(x₁, x₂)) = 2 + x₁ + x₂
POL(IF(x₁, x₂, x₃, x₄, x₅, x₆)) = 2·x₁ + x₅ + x₆
POL(IF2(x₁, x₂, x₃, x₄, x₅)) = 2·x₄ + x₅
POL(IF3(x₁, x₂, x₃, x₄)) = 2·x₃ + x₄
POL(IF4(x₁, x₂, x₃)) = 2·x₁ + x₃
POL(false) = 1
POL(le(x₁, x₂)) = 1 + x₁
POL(s(x₁)) = 2 + x₁
POL(true) = 1

(141) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AVERAGE(0, s(s(s(x0)))) → IF(true, false, false, false, 0, s(s(s(x0))))
IF2(false, false, false, 0, s(s(x1))) → IF3(false, false, 0, s(s(x1)))
IF3(false, false, 0, s(s(x1))) → IF4(false, 0, s(s(x1)))
AVERAGE(s(0), s(s(y_5))) → IF(false, false, false, le(y_5, 0), s(0), s(s(y_5)))
IF(false, false, false, y_0, s(0), s(s(z1))) → AVERAGE(0, s(s(s(z1))))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(142) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 5 less nodes.

(143) TRUE

(144) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, false, x0, s(s(y_0)), s(s(x2))) → AVERAGE(s(y_0), s(s(s(x2))))
AVERAGE(s(s(0)), s(s(x1))) → IF(false, false, false, le(x1, 0), s(s(0)), s(s(x1)))
AVERAGE(s(s(y_4)), s(s(x1))) → IF(false, false, false, le(x1, 0), s(s(y_4)), s(s(x1)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(145) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(146) Obligation:

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, false, x0, s(s(y_0)), s(s(x2))) → AVERAGE(s(y_0), s(s(s(x2))))
AVERAGE(s(s(0)), s(s(x1))) → IF(false, false, false, le(x1, 0), s(s(0)), s(s(x1)))
AVERAGE(s(s(y_4)), s(s(x1))) → IF(false, false, false, le(x1, 0), s(s(y_4)), s(s(x1)))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.

(147) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

IF(false, false, false, x0, s(s(y_0)), s(s(x2))) → AVERAGE(s(y_0), s(s(s(x2))))
The graph contains the following edges 5 > 1
AVERAGE(s(s(0)), s(s(x1))) → IF(false, false, false, le(x1, 0), s(s(0)), s(s(x1)))
The graph contains the following edges 1 >= 5, 2 >= 6
AVERAGE(s(s(y_4)), s(s(x1))) → IF(false, false, false, le(x1, 0), s(s(y_4)), s(s(x1)))
The graph contains the following edges 1 >= 5, 2 >= 6