Termination w.r.t. Q proof of AProVE_06_div

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 QDPSizeChangeProof (⇔, 0 ms)

        ↳20 YES

    ↳21 QDP

        ↳22 UsableRulesProof (⇔, 0 ms)

        ↳23 QDP

        ↳24 QReductionProof (⇔, 0 ms)

        ↳25 QDP

        ↳26 QDPSizeChangeProof (⇔, 0 ms)

        ↳27 YES

    ↳28 QDP

        ↳29 UsableRulesProof (⇔, 0 ms)

        ↳30 QDP

        ↳31 QReductionProof (⇔, 0 ms)

        ↳32 QDP

        ↳33 QDPSizeChangeProof (⇔, 0 ms)

        ↳34 YES

    ↳35 QDP

        ↳36 UsableRulesProof (⇔, 0 ms)

        ↳37 QDP

        ↳38 QReductionProof (⇔, 0 ms)

        ↳39 QDP

        ↳40 QDPSizeChangeProof (⇔, 0 ms)

        ↳41 YES

    ↳42 QDP

        ↳43 UsableRulesProof (⇔, 0 ms)

        ↳44 QDP

        ↳45 QReductionProof (⇔, 0 ms)

        ↳46 QDP

        ↳47 QDPSizeChangeProof (⇔, 0 ms)

        ↳48 YES

    ↳49 QDP

        ↳50 UsableRulesProof (⇔, 0 ms)

        ↳51 QDP

        ↳52 QReductionProof (⇔, 0 ms)

        ↳53 QDP

        ↳54 QDPSizeChangeProof (⇔, 0 ms)

        ↳55 YES

    ↳56 QDP

        ↳57 UsableRulesProof (⇔, 0 ms)

        ↳58 QDP

        ↳59 QReductionProof (⇔, 0 ms)

        ↳60 QDP

        ↳61 QDPSizeChangeProof (⇔, 0 ms)

        ↳62 YES

    ↳63 QDP

        ↳64 UsableRulesProof (⇔, 0 ms)

        ↳65 QDP

        ↳66 QReductionProof (⇔, 0 ms)

        ↳67 QDP

        ↳68 TransformationProof (⇔, 0 ms)

        ↳69 QDP

        ↳70 DependencyGraphProof (⇔, 0 ms)

        ↳71 QDP

        ↳72 TransformationProof (⇔, 0 ms)

        ↳73 QDP

        ↳74 DependencyGraphProof (⇔, 0 ms)

        ↳75 QDP

        ↳76 TransformationProof (⇔, 0 ms)

        ↳77 QDP

        ↳78 TransformationProof (⇔, 0 ms)

        ↳79 QDP

        ↳80 TransformationProof (⇔, 0 ms)

        ↳81 QDP

        ↳82 TransformationProof (⇔, 0 ms)

        ↳83 QDP

        ↳84 DependencyGraphProof (⇔, 0 ms)

        ↳85 QDP

        ↳86 TransformationProof (⇔, 0 ms)

        ↳87 QDP

        ↳88 TransformationProof (⇔, 0 ms)

        ↳89 QDP

        ↳90 TransformationProof (⇔, 0 ms)

        ↳91 QDP

        ↳92 TransformationProof (⇔, 0 ms)

        ↳93 QDP

        ↳94 DependencyGraphProof (⇔, 0 ms)

        ↳95 AND

            ↳96 QDP

                ↳97 TransformationProof (⇔, 0 ms)

                ↳98 QDP

                ↳99 TransformationProof (⇔, 0 ms)

                ↳100 QDP

                ↳101 QDPOrderProof (⇔, 0 ms)

                ↳102 QDP

                ↳103 DependencyGraphProof (⇔, 0 ms)

                ↳104 TRUE

            ↳105 QDP

                ↳106 UsableRulesProof (⇔, 0 ms)

                ↳107 QDP

                ↳108 TransformationProof (⇔, 0 ms)

                ↳109 QDP

                ↳110 TransformationProof (⇔, 0 ms)

                ↳111 QDP

                ↳112 MRRProof (⇔, 0 ms)

                ↳113 QDP

                ↳114 DependencyGraphProof (⇔, 0 ms)

                ↳115 TRUE

(0) Obligation:

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), 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:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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:

GE(s(x), 0) → GE(x, 0)
GE(0, s(s(x))) → GE(0, s(x))
GE(s(x), s(y)) → GE(x, y)
MINUS(0, s(x)) → MINUS(0, x)
MINUS(s(x), 0) → MINUS(x, 0)
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(0, s(x)) → PLUS(0, x)
PLUS(s(x), y) → PLUS(x, y)
DIV(x, y) → IFY(ge(y, s(0)), x, y)
DIV(x, y) → GE(y, s(0))
IFY(true, x, y) → IF(ge(x, y), x, y)
IFY(true, x, y) → GE(x, y)
IF(true, x, y) → DIV(minus(x, y), y)
IF(true, x, y) → MINUS(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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 9 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(0, s(x)) → PLUS(0, x)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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:

PLUS(0, s(x)) → PLUS(0, x)

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

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(11) Obligation:

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

PLUS(0, s(x)) → PLUS(0, x)

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:

PLUS(0, s(x)) → PLUS(0, x)
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:

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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:

PLUS(s(x), y) → PLUS(x, y)

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

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, 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].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(18) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

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

(19) 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:

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

(20) YES

(21) Obligation:

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

MINUS(s(x), 0) → MINUS(x, 0)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(22) 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.

(23) Obligation:

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

MINUS(s(x), 0) → MINUS(x, 0)

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

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(24) 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].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(25) Obligation:

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

MINUS(s(x), 0) → MINUS(x, 0)

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

(26) 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:

MINUS(s(x), 0) → MINUS(x, 0)
The graph contains the following edges 1 > 1, 2 >= 2

(27) YES

(28) Obligation:

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

MINUS(0, s(x)) → MINUS(0, x)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(29) 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.

(30) Obligation:

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

MINUS(0, s(x)) → MINUS(0, x)

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

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(31) 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].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(32) Obligation:

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

MINUS(0, s(x)) → MINUS(0, x)

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

(33) 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:

MINUS(0, s(x)) → MINUS(0, x)
The graph contains the following edges 1 >= 1, 2 > 2

(34) YES

(35) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(36) 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.

(37) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

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

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(38) 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].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(39) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

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

(40) 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:

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

(41) YES

(42) Obligation:

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

GE(0, s(s(x))) → GE(0, s(x))

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(43) 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.

(44) Obligation:

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

GE(0, s(s(x))) → GE(0, s(x))

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

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(45) 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].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(46) Obligation:

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

GE(0, s(s(x))) → GE(0, s(x))

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

(47) 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:

GE(0, s(s(x))) → GE(0, s(x))
The graph contains the following edges 1 >= 1, 2 > 2

(48) YES

(49) Obligation:

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

GE(s(x), 0) → GE(x, 0)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(50) 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.

(51) Obligation:

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

GE(s(x), 0) → GE(x, 0)

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

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(52) 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].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(53) Obligation:

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

GE(s(x), 0) → GE(x, 0)

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

(54) 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:

GE(s(x), 0) → GE(x, 0)
The graph contains the following edges 1 > 1, 2 >= 2

(55) YES

(56) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(57) 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.

(58) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

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

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(59) 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].

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(60) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

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

(61) 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:

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

(62) YES

(63) Obligation:

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

DIV(x, y) → IFY(ge(y, s(0)), x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(64) 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.

(65) Obligation:

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

DIV(x, y) → IFY(ge(y, s(0)), x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

(66) 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].

plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

(67) Obligation:

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

DIV(x, y) → IFY(ge(y, s(0)), x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(68) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule DIV(x, y) → IFY(ge(y, s(0)), x, y) at position [0] we obtained the following new rules [LPAR04]:

DIV(y0, 0) → IFY(false, y0, 0) → DIV(y0, 0) → IFY(false, y0, 0)
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0)) → DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))

(69) Obligation:

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

IFY(true, x, y) → IF(ge(x, y), x, y)
IF(true, x, y) → DIV(minus(x, y), y)
DIV(y0, 0) → IFY(false, y0, 0)
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(70) DependencyGraphProof (EQUIVALENT transformation)

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

(71) Obligation:

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

IF(true, x, y) → DIV(minus(x, y), y)
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IFY(true, x, y) → IF(ge(x, y), x, y)

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(72) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IFY(true, x, y) → IF(ge(x, y), x, y) at position [0] we obtained the following new rules [LPAR04]:

IFY(true, 0, 0) → IF(true, 0, 0) → IFY(true, 0, 0) → IF(true, 0, 0)
IFY(true, s(x0), 0) → IF(ge(x0, 0), s(x0), 0) → IFY(true, s(x0), 0) → IF(ge(x0, 0), s(x0), 0)
IFY(true, 0, s(0)) → IF(false, 0, s(0)) → IFY(true, 0, s(0)) → IF(false, 0, s(0))
IFY(true, 0, s(s(x0))) → IF(ge(0, s(x0)), 0, s(s(x0))) → IFY(true, 0, s(s(x0))) → IF(ge(0, s(x0)), 0, s(s(x0)))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1)) → IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))

(73) Obligation:

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

IF(true, x, y) → DIV(minus(x, y), y)
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IFY(true, 0, 0) → IF(true, 0, 0)
IFY(true, s(x0), 0) → IF(ge(x0, 0), s(x0), 0)
IFY(true, 0, s(0)) → IF(false, 0, s(0))
IFY(true, 0, s(s(x0))) → IF(ge(0, s(x0)), 0, s(s(x0)))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(74) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(75) Obligation:

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

DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(76) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(true, x, y) → DIV(minus(x, y), y) we obtained the following new rules [LPAR04]:

IF(true, s(z0), s(z1)) → DIV(minus(s(z0), s(z1)), s(z1)) → IF(true, s(z0), s(z1)) → DIV(minus(s(z0), s(z1)), s(z1))

(77) Obligation:

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

DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(z0), s(z1)) → DIV(minus(s(z0), s(z1)), s(z1))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(78) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF(true, s(z0), s(z1)) → DIV(minus(s(z0), s(z1)), s(z1)) at position [0] we obtained the following new rules [LPAR04]:

IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1)) → IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1))

(79) Obligation:

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

DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(80) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0)) we obtained the following new rules [LPAR04]:

DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1)) → DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1))

(81) Obligation:

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1))
DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(82) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1)) at position [0] we obtained the following new rules [LPAR04]:

IF(true, s(0), s(0)) → DIV(0, s(0)) → IF(true, s(0), s(0)) → DIV(0, s(0))
IF(true, s(0), s(s(x0))) → DIV(minus(0, x0), s(s(x0))) → IF(true, s(0), s(s(x0))) → DIV(minus(0, x0), s(s(x0)))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0)) → IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1))) → IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

(83) Obligation:

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1))
IF(true, s(0), s(0)) → DIV(0, s(0))
IF(true, s(0), s(s(x0))) → DIV(minus(0, x0), s(s(x0)))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(84) DependencyGraphProof (EQUIVALENT transformation)

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

(85) Obligation:

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

IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(86) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1)) we obtained the following new rules [LPAR04]:

DIV(s(y_0), s(0)) → IFY(ge(0, 0), s(y_0), s(0)) → DIV(s(y_0), s(0)) → IFY(ge(0, 0), s(y_0), s(0))
DIV(s(x0), s(s(z1))) → IFY(ge(s(z1), 0), s(x0), s(s(z1))) → DIV(s(x0), s(s(z1))) → IFY(ge(s(z1), 0), s(x0), s(s(z1)))

(87) Obligation:

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

IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
DIV(s(y_0), s(0)) → IFY(ge(0, 0), s(y_0), s(0))
DIV(s(x0), s(s(z1))) → IFY(ge(s(z1), 0), s(x0), s(s(z1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(88) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule DIV(s(y_0), s(0)) → IFY(ge(0, 0), s(y_0), s(0)) at position [0] we obtained the following new rules [LPAR04]:

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0)) → DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))

(89) Obligation:

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

IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
DIV(s(x0), s(s(z1))) → IFY(ge(s(z1), 0), s(x0), s(s(z1)))
DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(90) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule DIV(s(x0), s(s(z1))) → IFY(ge(s(z1), 0), s(x0), s(s(z1))) at position [0] we obtained the following new rules [LPAR04]:

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1))) → DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))

(91) Obligation:

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

IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(92) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1)) we obtained the following new rules [LPAR04]:

IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0)) → IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1))) → IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1)))

(93) Obligation:

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

IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(94) DependencyGraphProof (EQUIVALENT transformation)

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

(95) Complex Obligation (AND)

(96) Obligation:

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

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(97) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1))) we obtained the following new rules [LPAR04]:

IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(s(y_1), s(x1)), s(s(y_1)), s(s(x1))) → IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(s(y_1), s(x1)), s(s(y_1)), s(s(x1)))

(98) Obligation:

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

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(s(y_1), s(x1)), s(s(y_1)), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(99) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(s(y_1), s(x1)), s(s(y_1)), s(s(x1))) at position [0] we obtained the following new rules [LPAR04]:

IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1))) → IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1)))

(100) Obligation:

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

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(101) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( IF(x₁, ..., x₃) ) = max{0, 2x₁ + 2x₂ + x₃ - 1}

POL( IFY(x₁, ..., x₃) ) = max{0, 2x₁ + 2x₂ + x₃ - 1}

POL( ge(x₁, x₂) ) = 1

POL( 0 ) = 0

POL( true ) = 1

POL( s(x₁) ) = 2x₁ + 1

POL( DIV(x₁, x₂) ) = 2x₁ + x₂ + 2

POL( minus(x₁, x₂) ) = 2x₁ + 1

POL( false ) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

(102) Obligation:

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

IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(103) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(104) TRUE

(105) Obligation:

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

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(106) 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.

(107) Obligation:

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

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(s(x), 0) → s(minus(x, 0))
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(108) TransformationProof (EQUIVALENT transformation)

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

IFY(true, s(s(y_1)), s(0)) → IF(ge(s(y_1), 0), s(s(y_1)), s(0)) → IFY(true, s(s(y_1)), s(0)) → IF(ge(s(y_1), 0), s(s(y_1)), s(0))

(109) Obligation:

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

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IFY(true, s(s(y_1)), s(0)) → IF(ge(s(y_1), 0), s(s(y_1)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(s(x), 0) → s(minus(x, 0))
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(110) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IFY(true, s(s(y_1)), s(0)) → IF(ge(s(y_1), 0), s(s(y_1)), s(0)) at position [0] we obtained the following new rules [LPAR04]:

IFY(true, s(s(y_1)), s(0)) → IF(ge(y_1, 0), s(s(y_1)), s(0)) → IFY(true, s(s(y_1)), s(0)) → IF(ge(y_1, 0), s(s(y_1)), s(0))

(111) Obligation:

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

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IFY(true, s(s(y_1)), s(0)) → IF(ge(y_1, 0), s(s(y_1)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(s(x), 0) → s(minus(x, 0))
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(112) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IFY(true, s(s(y_1)), s(0)) → IF(ge(y_1, 0), s(s(y_1)), s(0))

Strictly oriented rules of the TRS R:

ge(s(x), 0) → ge(x, 0)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0
POL(DIV(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(IF(x₁, x₂, x₃)) = 2·x₁ + x₂ + x₃
POL(IFY(x₁, x₂, x₃)) = 1 + x₁ + 2·x₂ + x₃
POL(ge(x₁, x₂)) = 2·x₁ + 2·x₂
POL(minus(x₁, x₂)) = x₁ + 2·x₂
POL(s(x₁)) = 2 + 2·x₁
POL(true) = 0

(113) Obligation:

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

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(s(x), 0) → s(minus(x, 0))
ge(0, 0) → true

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

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

(114) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.