Termination w.r.t. Q proof of Beerendonk_07

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

0 QTRS

↳1 AAECC Innermost (⇔, 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 TransformationProof (⇔, 0 ms)

        ↳27 QDP

        ↳28 TransformationProof (⇔, 0 ms)

        ↳29 QDP

        ↳30 TransformationProof (⇔, 0 ms)

        ↳31 QDP

        ↳32 TransformationProof (⇔, 0 ms)

        ↳33 QDP

        ↳34 DependencyGraphProof (⇔, 0 ms)

        ↳35 QDP

        ↳36 TransformationProof (⇔, 0 ms)

        ↳37 QDP

        ↳38 TransformationProof (⇔, 0 ms)

        ↳39 QDP

        ↳40 TransformationProof (⇔, 0 ms)

        ↳41 QDP

        ↳42 TransformationProof (⇔, 0 ms)

        ↳43 QDP

        ↳44 DependencyGraphProof (⇔, 0 ms)

        ↳45 QDP

        ↳46 TransformationProof (⇔, 0 ms)

        ↳47 QDP

        ↳48 DependencyGraphProof (⇔, 0 ms)

        ↳49 QDP

        ↳50 TransformationProof (⇔, 0 ms)

        ↳51 QDP

        ↳52 DependencyGraphProof (⇔, 0 ms)

        ↳53 AND

            ↳54 QDP

                ↳55 UsableRulesProof (⇔, 0 ms)

                ↳56 QDP

                ↳57 QReductionProof (⇔, 0 ms)

                ↳58 QDP

                ↳59 TransformationProof (⇔, 0 ms)

                ↳60 QDP

                ↳61 UsableRulesProof (⇔, 0 ms)

                ↳62 QDP

                ↳63 TransformationProof (⇔, 0 ms)

                ↳64 QDP

                ↳65 UsableRulesProof (⇔, 0 ms)

                ↳66 QDP

                ↳67 QReductionProof (⇔, 0 ms)

                ↳68 QDP

                ↳69 TransformationProof (⇔, 0 ms)

                ↳70 QDP

                ↳71 TransformationProof (⇔, 0 ms)

                ↳72 QDP

                ↳73 TransformationProof (⇔, 0 ms)

                ↳74 QDP

                ↳75 QDPSizeChangeProof (⇔, 0 ms)

                ↳76 YES

            ↳77 QDP

                ↳78 UsableRulesProof (⇔, 0 ms)

                ↳79 QDP

                ↳80 QReductionProof (⇔, 0 ms)

                ↳81 QDP

                ↳82 QDPSizeChangeProof (⇔, 0 ms)

                ↳83 YES

(0) Obligation:

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

cond1(true, x, y, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(gr(add(x, y), z), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(gr(add(x, y), z), x, p(y), z)
cond3(false, x, y, z) → cond1(gr(add(x, y), z), x, y, z)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
p(0) → 0
p(s(x)) → x

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond1(true, x, y, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(gr(add(x, y), z), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(gr(add(x, y), z), x, p(y), z)
cond3(false, x, y, z) → cond1(gr(add(x, y), z), x, y, z)

The signature Sigma is {cond1, cond2, cond3}

(2) Obligation:

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

cond1(true, x, y, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(gr(add(x, y), z), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(gr(add(x, y), z), x, p(y), z)
cond3(false, x, y, z) → cond1(gr(add(x, y), z), x, y, z)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

(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:

COND1(true, x, y, z) → COND2(gr(x, 0), x, y, z)
COND1(true, x, y, z) → GR(x, 0)
COND2(true, x, y, z) → COND1(gr(add(x, y), z), p(x), y, z)
COND2(true, x, y, z) → GR(add(x, y), z)
COND2(true, x, y, z) → ADD(x, y)
COND2(true, x, y, z) → P(x)
COND2(false, x, y, z) → COND3(gr(y, 0), x, y, z)
COND2(false, x, y, z) → GR(y, 0)
COND3(true, x, y, z) → COND1(gr(add(x, y), z), x, p(y), z)
COND3(true, x, y, z) → GR(add(x, y), z)
COND3(true, x, y, z) → ADD(x, y)
COND3(true, x, y, z) → P(y)
COND3(false, x, y, z) → COND1(gr(add(x, y), z), x, y, z)
COND3(false, x, y, z) → GR(add(x, y), z)
COND3(false, x, y, z) → ADD(x, y)
GR(s(x), s(y)) → GR(x, y)
ADD(s(x), y) → ADD(x, y)

The TRS R consists of the following rules:

cond1(true, x, y, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(gr(add(x, y), z), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(gr(add(x, y), z), x, p(y), z)
cond3(false, x, y, z) → cond1(gr(add(x, y), z), x, y, z)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 10 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

ADD(s(x), y) → ADD(x, y)

The TRS R consists of the following rules:

cond1(true, x, y, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(gr(add(x, y), z), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(gr(add(x, y), z), x, p(y), z)
cond3(false, x, y, z) → cond1(gr(add(x, y), z), x, y, z)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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:

ADD(s(x), y) → ADD(x, y)

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

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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].

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

(11) Obligation:

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

ADD(s(x), y) → ADD(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:

ADD(s(x), y) → ADD(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:

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(gr(add(x, y), z), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(gr(add(x, y), z), x, p(y), z)
cond3(false, x, y, z) → cond1(gr(add(x, y), z), x, y, z)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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:

GR(s(x), s(y)) → GR(x, y)

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

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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].

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

(18) Obligation:

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

GR(s(x), s(y)) → GR(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:

GR(s(x), s(y)) → GR(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:

COND2(true, x, y, z) → COND1(gr(add(x, y), z), p(x), y, z)
COND1(true, x, y, z) → COND2(gr(x, 0), x, y, z)
COND2(false, x, y, z) → COND3(gr(y, 0), x, y, z)
COND3(true, x, y, z) → COND1(gr(add(x, y), z), x, p(y), z)
COND3(false, x, y, z) → COND1(gr(add(x, y), z), x, y, z)

The TRS R consists of the following rules:

cond1(true, x, y, z) → cond2(gr(x, 0), x, y, z)
cond2(true, x, y, z) → cond1(gr(add(x, y), z), p(x), y, z)
cond2(false, x, y, z) → cond3(gr(y, 0), x, y, z)
cond3(true, x, y, z) → cond1(gr(add(x, y), z), x, p(y), z)
cond3(false, x, y, z) → cond1(gr(add(x, y), z), x, y, z)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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:

COND2(true, x, y, z) → COND1(gr(add(x, y), z), p(x), y, z)
COND1(true, x, y, z) → COND2(gr(x, 0), x, y, z)
COND2(false, x, y, z) → COND3(gr(y, 0), x, y, z)
COND3(true, x, y, z) → COND1(gr(add(x, y), z), x, p(y), z)
COND3(false, x, y, z) → COND1(gr(add(x, y), z), x, y, z)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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].

cond1(true, x0, x1, x2)
cond2(true, x0, x1, x2)
cond2(false, x0, x1, x2)
cond3(true, x0, x1, x2)
cond3(false, x0, x1, x2)

(25) Obligation:

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

COND2(true, x, y, z) → COND1(gr(add(x, y), z), p(x), y, z)
COND1(true, x, y, z) → COND2(gr(x, 0), x, y, z)
COND2(false, x, y, z) → COND3(gr(y, 0), x, y, z)
COND3(true, x, y, z) → COND1(gr(add(x, y), z), x, p(y), z)
COND3(false, x, y, z) → COND1(gr(add(x, y), z), x, y, z)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(26) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND2(true, x, y, z) → COND1(gr(add(x, y), z), p(x), y, z) at position [0] we obtained the following new rules [LPAR04]:

COND2(true, 0, x0, y2) → COND1(gr(x0, y2), p(0), x0, y2) → COND2(true, 0, x0, y2) → COND1(gr(x0, y2), p(0), x0, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), p(s(x0)), x1, y2) → COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), p(s(x0)), x1, y2)

(27) Obligation:

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

COND1(true, x, y, z) → COND2(gr(x, 0), x, y, z)
COND2(false, x, y, z) → COND3(gr(y, 0), x, y, z)
COND3(true, x, y, z) → COND1(gr(add(x, y), z), x, p(y), z)
COND3(false, x, y, z) → COND1(gr(add(x, y), z), x, y, z)
COND2(true, 0, x0, y2) → COND1(gr(x0, y2), p(0), x0, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), p(s(x0)), x1, y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(28) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, 0, x0, y2) → COND1(gr(x0, y2), p(0), x0, y2) at position [1] we obtained the following new rules [LPAR04]:

COND2(true, 0, x0, y2) → COND1(gr(x0, y2), 0, x0, y2) → COND2(true, 0, x0, y2) → COND1(gr(x0, y2), 0, x0, y2)

(29) Obligation:

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

COND1(true, x, y, z) → COND2(gr(x, 0), x, y, z)
COND2(false, x, y, z) → COND3(gr(y, 0), x, y, z)
COND3(true, x, y, z) → COND1(gr(add(x, y), z), x, p(y), z)
COND3(false, x, y, z) → COND1(gr(add(x, y), z), x, y, z)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), p(s(x0)), x1, y2)
COND2(true, 0, x0, y2) → COND1(gr(x0, y2), 0, x0, y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(30) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), p(s(x0)), x1, y2) at position [1] we obtained the following new rules [LPAR04]:

COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2) → COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)

(31) Obligation:

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

COND1(true, x, y, z) → COND2(gr(x, 0), x, y, z)
COND2(false, x, y, z) → COND3(gr(y, 0), x, y, z)
COND3(true, x, y, z) → COND1(gr(add(x, y), z), x, p(y), z)
COND3(false, x, y, z) → COND1(gr(add(x, y), z), x, y, z)
COND2(true, 0, x0, y2) → COND1(gr(x0, y2), 0, x0, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(32) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND1(true, x, y, z) → COND2(gr(x, 0), x, y, z) at position [0] we obtained the following new rules [LPAR04]:

COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2) → COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2) → COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)

(33) Obligation:

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

COND2(false, x, y, z) → COND3(gr(y, 0), x, y, z)
COND3(true, x, y, z) → COND1(gr(add(x, y), z), x, p(y), z)
COND3(false, x, y, z) → COND1(gr(add(x, y), z), x, y, z)
COND2(true, 0, x0, y2) → COND1(gr(x0, y2), 0, x0, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(34) DependencyGraphProof (EQUIVALENT transformation)

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

(35) Obligation:

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

COND3(true, x, y, z) → COND1(gr(add(x, y), z), x, p(y), z)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND2(false, x, y, z) → COND3(gr(y, 0), x, y, z)
COND3(false, x, y, z) → COND1(gr(add(x, y), z), x, y, z)
COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(36) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(true, x, y, z) → COND1(gr(add(x, y), z), x, p(y), z) at position [0] we obtained the following new rules [LPAR04]:

COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2) → COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2)
COND3(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), p(x1), y2) → COND3(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), p(x1), y2)

(37) Obligation:

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

COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND2(false, x, y, z) → COND3(gr(y, 0), x, y, z)
COND3(false, x, y, z) → COND1(gr(add(x, y), z), x, y, z)
COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)
COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2)
COND3(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), p(x1), y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(38) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND2(false, x, y, z) → COND3(gr(y, 0), x, y, z) at position [0] we obtained the following new rules [LPAR04]:

COND2(false, y0, 0, y2) → COND3(false, y0, 0, y2) → COND2(false, y0, 0, y2) → COND3(false, y0, 0, y2)
COND2(false, y0, s(x0), y2) → COND3(true, y0, s(x0), y2) → COND2(false, y0, s(x0), y2) → COND3(true, y0, s(x0), y2)

(39) Obligation:

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

COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND3(false, x, y, z) → COND1(gr(add(x, y), z), x, y, z)
COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)
COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2)
COND3(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), p(x1), y2)
COND2(false, y0, 0, y2) → COND3(false, y0, 0, y2)
COND2(false, y0, s(x0), y2) → COND3(true, y0, s(x0), y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(40) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(false, x, y, z) → COND1(gr(add(x, y), z), x, y, z) at position [0] we obtained the following new rules [LPAR04]:

COND3(false, 0, x0, y2) → COND1(gr(x0, y2), 0, x0, y2) → COND3(false, 0, x0, y2) → COND1(gr(x0, y2), 0, x0, y2)
COND3(false, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), x1, y2) → COND3(false, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), x1, y2)

(41) Obligation:

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

COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)
COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2)
COND3(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), p(x1), y2)
COND2(false, y0, 0, y2) → COND3(false, y0, 0, y2)
COND2(false, y0, s(x0), y2) → COND3(true, y0, s(x0), y2)
COND3(false, 0, x0, y2) → COND1(gr(x0, y2), 0, x0, y2)
COND3(false, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), x1, y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(42) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND3(false, 0, x0, y2) → COND1(gr(x0, y2), 0, x0, y2) at position [0] we obtained the following new rules [LPAR04]:

COND3(false, 0, 0, x0) → COND1(false, 0, 0, x0) → COND3(false, 0, 0, x0) → COND1(false, 0, 0, x0)
COND3(false, 0, s(x0), 0) → COND1(true, 0, s(x0), 0) → COND3(false, 0, s(x0), 0) → COND1(true, 0, s(x0), 0)
COND3(false, 0, s(x0), s(x1)) → COND1(gr(x0, x1), 0, s(x0), s(x1)) → COND3(false, 0, s(x0), s(x1)) → COND1(gr(x0, x1), 0, s(x0), s(x1))

(43) Obligation:

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

COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)
COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2)
COND3(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), p(x1), y2)
COND2(false, y0, 0, y2) → COND3(false, y0, 0, y2)
COND2(false, y0, s(x0), y2) → COND3(true, y0, s(x0), y2)
COND3(false, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), x1, y2)
COND3(false, 0, 0, x0) → COND1(false, 0, 0, x0)
COND3(false, 0, s(x0), 0) → COND1(true, 0, s(x0), 0)
COND3(false, 0, s(x0), s(x1)) → COND1(gr(x0, x1), 0, s(x0), s(x1))

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(44) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(45) Obligation:

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

COND2(false, y0, 0, y2) → COND3(false, y0, 0, y2)
COND3(false, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), x1, y2)
COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND2(false, y0, s(x0), y2) → COND3(true, y0, s(x0), y2)
COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2)
COND3(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), p(x1), y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(46) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND2(false, y0, 0, y2) → COND3(false, y0, 0, y2) we obtained the following new rules [LPAR04]:

COND2(false, 0, 0, z1) → COND3(false, 0, 0, z1) → COND2(false, 0, 0, z1) → COND3(false, 0, 0, z1)

(47) Obligation:

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

COND3(false, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), x1, y2)
COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND2(false, y0, s(x0), y2) → COND3(true, y0, s(x0), y2)
COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2)
COND3(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), p(x1), y2)
COND2(false, 0, 0, z1) → COND3(false, 0, 0, z1)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(48) DependencyGraphProof (EQUIVALENT transformation)

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

(49) Obligation:

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

COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND2(false, y0, s(x0), y2) → COND3(true, y0, s(x0), y2)
COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2)
COND3(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), p(x1), y2)
COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(50) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND2(false, y0, s(x0), y2) → COND3(true, y0, s(x0), y2) we obtained the following new rules [LPAR04]:

COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1) → COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1)

(51) Obligation:

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

COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2)
COND3(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), s(x0), p(x1), y2)
COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)
COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(52) DependencyGraphProof (EQUIVALENT transformation)

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

(53) Complex Obligation (AND)

(54) Obligation:

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

COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1)
COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(55) 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.

(56) Obligation:

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

COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1)
COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(57) 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].

add(0, x0)
add(s(x0), x1)

(58) Obligation:

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

COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1)
COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(59) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND3(true, 0, x0, y2) → COND1(gr(x0, y2), 0, p(x0), y2) we obtained the following new rules [LPAR04]:

COND3(true, 0, s(z0), z1) → COND1(gr(s(z0), z1), 0, p(s(z0)), z1) → COND3(true, 0, s(z0), z1) → COND1(gr(s(z0), z1), 0, p(s(z0)), z1)

(60) Obligation:

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

COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND3(true, 0, s(z0), z1) → COND1(gr(s(z0), z1), 0, p(s(z0)), z1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(61) 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.

(62) Obligation:

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

COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND3(true, 0, s(z0), z1) → COND1(gr(s(z0), z1), 0, p(s(z0)), z1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(63) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND3(true, 0, s(z0), z1) → COND1(gr(s(z0), z1), 0, p(s(z0)), z1) at position [2] we obtained the following new rules [LPAR04]:

COND3(true, 0, s(z0), z1) → COND1(gr(s(z0), z1), 0, z0, z1) → COND3(true, 0, s(z0), z1) → COND1(gr(s(z0), z1), 0, z0, z1)

(64) Obligation:

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

COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND3(true, 0, s(z0), z1) → COND1(gr(s(z0), z1), 0, z0, z1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(65) 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.

(66) Obligation:

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

COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND3(true, 0, s(z0), z1) → COND1(gr(s(z0), z1), 0, z0, z1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(67) 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(0)
p(s(x0))

(68) Obligation:

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

COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1)
COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2)
COND3(true, 0, s(z0), z1) → COND1(gr(s(z0), z1), 0, z0, z1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

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

(69) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule COND1(true, 0, y1, y2) → COND2(false, 0, y1, y2) we obtained the following new rules [LPAR04]:

COND1(true, 0, s(y_0), x1) → COND2(false, 0, s(y_0), x1) → COND1(true, 0, s(y_0), x1) → COND2(false, 0, s(y_0), x1)

(70) Obligation:

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

COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1)
COND3(true, 0, s(z0), z1) → COND1(gr(s(z0), z1), 0, z0, z1)
COND1(true, 0, s(y_0), x1) → COND2(false, 0, s(y_0), x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

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

(71) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule COND3(true, 0, s(z0), z1) → COND1(gr(s(z0), z1), 0, z0, z1) we obtained the following new rules [LPAR04]:

COND3(true, 0, s(s(y_1)), x1) → COND1(gr(s(s(y_1)), x1), 0, s(y_1), x1) → COND3(true, 0, s(s(y_1)), x1) → COND1(gr(s(s(y_1)), x1), 0, s(y_1), x1)

(72) Obligation:

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

COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1)
COND1(true, 0, s(y_0), x1) → COND2(false, 0, s(y_0), x1)
COND3(true, 0, s(s(y_1)), x1) → COND1(gr(s(s(y_1)), x1), 0, s(y_1), x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

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

(73) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule COND2(false, 0, s(x1), z1) → COND3(true, 0, s(x1), z1) we obtained the following new rules [LPAR04]:

COND2(false, 0, s(s(y_0)), x1) → COND3(true, 0, s(s(y_0)), x1) → COND2(false, 0, s(s(y_0)), x1) → COND3(true, 0, s(s(y_0)), x1)

(74) Obligation:

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

COND1(true, 0, s(y_0), x1) → COND2(false, 0, s(y_0), x1)
COND3(true, 0, s(s(y_1)), x1) → COND1(gr(s(s(y_1)), x1), 0, s(y_1), x1)
COND2(false, 0, s(s(y_0)), x1) → COND3(true, 0, s(s(y_0)), x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

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

(75) 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:

COND2(false, 0, s(s(y_0)), x1) → COND3(true, 0, s(s(y_0)), x1)
The graph contains the following edges 2 >= 2, 3 >= 3, 4 >= 4
COND3(true, 0, s(s(y_1)), x1) → COND1(gr(s(s(y_1)), x1), 0, s(y_1), x1)
The graph contains the following edges 2 >= 2, 3 > 3, 4 >= 4
COND1(true, 0, s(y_0), x1) → COND2(false, 0, s(y_0), x1)
The graph contains the following edges 2 >= 2, 3 >= 3, 4 >= 4

(76) YES

(77) Obligation:

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

COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(78) 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.

(79) Obligation:

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

COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)
p(0)
p(s(x0))

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

(80) 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(0)
p(s(x0))

(81) Obligation:

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

COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)
COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)

The TRS R consists of the following rules:

add(0, x) → x
add(s(x), y) → s(add(x, y))
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)

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

(82) 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:

COND2(true, s(x0), x1, y2) → COND1(gr(s(add(x0, x1)), y2), x0, x1, y2)
The graph contains the following edges 2 > 2, 3 >= 3, 4 >= 4
COND1(true, s(x0), y1, y2) → COND2(true, s(x0), y1, y2)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3, 4 >= 4