Termination w.r.t. Q proof of /tmp/tmpyCNw2N/divhard.trs

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 Overlay + Local Confluence (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

    ↳7 QDP

        ↳8 UsableRulesProof (⇔)

        ↳9 QDP

        ↳10 QReductionProof (⇔)

        ↳11 QDP

        ↳12 QDPSizeChangeProof (⇔)

        ↳13 YES

    ↳14 QDP

        ↳15 UsableRulesProof (⇔)

        ↳16 QDP

        ↳17 QReductionProof (⇔)

        ↳18 QDP

        ↳19 QDPSizeChangeProof (⇔)

        ↳20 YES

    ↳21 QDP

        ↳22 UsableRulesProof (⇔)

        ↳23 QDP

        ↳24 QReductionProof (⇔)

        ↳25 QDP

        ↳26 QDPOrderProof (⇔)

        ↳27 QDP

        ↳28 PisEmptyProof (⇔)

        ↳29 YES

    ↳30 QDP

        ↳31 UsableRulesProof (⇔)

        ↳32 QDP

        ↳33 QReductionProof (⇔)

        ↳34 QDP

        ↳35 Rewriting (⇔)

        ↳36 QDP

        ↳37 Narrowing (⇔)

        ↳38 QDP

        ↳39 DependencyGraphProof (⇔)

        ↳40 QDP

        ↳41 Narrowing (⇔)

        ↳42 QDP

        ↳43 DependencyGraphProof (⇔)

        ↳44 QDP

        ↳45 Narrowing (⇔)

        ↳46 QDP

        ↳47 DependencyGraphProof (⇔)

        ↳48 QDP

        ↳49 Instantiation (⇔)

        ↳50 QDP

            ↳51 MNOCProof (⇔)

                ↳52 QDP

            ↳53 MNOCProof (⇔)

                ↳54 QDP

(0) Obligation:

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

minus(x, y) → if(gt(x, y), x, y)
if(true, x, y) → s(minus(p(x), y))
if(false, x, y) → 0
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
div(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → 0
if2(true, x, y) → s(div(minus(x, y), y))
if2(false, x, y) → 0

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:

minus(x, y) → if(gt(x, y), x, y)
if(true, x, y) → s(minus(p(x), y))
if(false, x, y) → 0
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
div(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → 0
if2(true, x, y) → s(div(minus(x, y), y))
if2(false, x, y) → 0

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

MINUS(x, y) → IF(gt(x, y), x, y)
MINUS(x, y) → GT(x, y)
IF(true, x, y) → MINUS(p(x), y)
IF(true, x, y) → P(x)
GE(s(x), s(y)) → GE(x, y)
GT(s(x), s(y)) → GT(x, y)
DIV(x, y) → IF1(ge(x, y), x, y)
DIV(x, y) → GE(x, y)
IF1(true, x, y) → IF2(gt(y, 0), x, y)
IF1(true, x, y) → GT(y, 0)
IF2(true, x, y) → DIV(minus(x, y), y)
IF2(true, x, y) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, y) → if(gt(x, y), x, y)
if(true, x, y) → s(minus(p(x), y))
if(false, x, y) → 0
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
div(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → 0
if2(true, x, y) → s(div(minus(x, y), y))
if2(false, x, y) → 0

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

minus(x, y) → if(gt(x, y), x, y)
if(true, x, y) → s(minus(p(x), y))
if(false, x, y) → 0
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
div(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → 0
if2(true, x, y) → s(div(minus(x, y), y))
if2(false, x, y) → 0

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

GT(s(x), s(y)) → GT(x, y)

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

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

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

(10) QReductionProof (EQUIVALENT transformation)

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

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

(11) Obligation:

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

GT(s(x), s(y)) → GT(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:

GT(s(x), s(y)) → GT(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:

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

minus(x, y) → if(gt(x, y), x, y)
if(true, x, y) → s(minus(p(x), y))
if(false, x, y) → 0
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
div(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → 0
if2(true, x, y) → s(div(minus(x, y), y))
if2(false, x, y) → 0

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

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

(15) UsableRulesProof (EQUIVALENT transformation)

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

(16) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

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

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

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

(17) QReductionProof (EQUIVALENT transformation)

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

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

(18) 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.

(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:

GE(s(x), s(y)) → GE(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:

IF(true, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(gt(x, y), x, y)

The TRS R consists of the following rules:

minus(x, y) → if(gt(x, y), x, y)
if(true, x, y) → s(minus(p(x), y))
if(false, x, y) → 0
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
div(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → 0
if2(true, x, y) → s(div(minus(x, y), y))
if2(false, x, y) → 0

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, 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:

IF(true, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(gt(x, y), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, 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].

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

(25) Obligation:

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

IF(true, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(gt(x, y), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(true, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(gt(x, y), x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(0) = 0
POL(IF(x₁, x₂, x₃)) = [1/4] + x₁ + x₂
POL(MINUS(x₁, x₂)) = [2] + [2]x₁
POL(false) = [1/4]
POL(gt(x₁, x₂)) = [1] + x₁
POL(p(x₁)) = [1/4]x₁
POL(s(x₁)) = [4] + [4]x₁
POL(true) = [4]

The value of delta used in the strict ordering is 3/4.
The following usable rules [FROCOS05] were oriented:

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

(27) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(28) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(29) YES

(30) Obligation:

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

IF2(true, x, y) → DIV(minus(x, y), y)
DIV(x, y) → IF1(ge(x, y), x, y)
IF1(true, x, y) → IF2(gt(y, 0), x, y)

The TRS R consists of the following rules:

minus(x, y) → if(gt(x, y), x, y)
if(true, x, y) → s(minus(p(x), y))
if(false, x, y) → 0
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
div(x, y) → if1(ge(x, y), x, y)
if1(true, x, y) → if2(gt(y, 0), x, y)
if1(false, x, y) → 0
if2(true, x, y) → s(div(minus(x, y), y))
if2(false, x, y) → 0

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

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

(31) 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.

(32) Obligation:

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

IF2(true, x, y) → DIV(minus(x, y), y)
DIV(x, y) → IF1(ge(x, y), x, y)
IF1(true, x, y) → IF2(gt(y, 0), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

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

(33) 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].

div(x0, x1)
if1(true, x0, x1)
if1(false, x0, x1)
if2(true, x0, x1)
if2(false, x0, x1)

(34) Obligation:

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

IF2(true, x, y) → DIV(minus(x, y), y)
DIV(x, y) → IF1(ge(x, y), x, y)
IF1(true, x, y) → IF2(gt(y, 0), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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

(35) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF2(true, x, y) → DIV(minus(x, y), y) at position [0] we obtained the following new rules [LPAR04]:

IF2(true, x, y) → DIV(if(gt(x, y), x, y), y)

(36) Obligation:

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

DIV(x, y) → IF1(ge(x, y), x, y)
IF1(true, x, y) → IF2(gt(y, 0), x, y)
IF2(true, x, y) → DIV(if(gt(x, y), x, y), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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

(37) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule DIV(x, y) → IF1(ge(x, y), x, y) at position [0] we obtained the following new rules [LPAR04]:

DIV(x0, 0) → IF1(true, x0, 0)
DIV(0, s(x0)) → IF1(false, 0, s(x0))
DIV(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))

(38) Obligation:

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

IF1(true, x, y) → IF2(gt(y, 0), x, y)
IF2(true, x, y) → DIV(if(gt(x, y), x, y), y)
DIV(x0, 0) → IF1(true, x0, 0)
DIV(0, s(x0)) → IF1(false, 0, s(x0))
DIV(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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

(39) DependencyGraphProof (EQUIVALENT transformation)

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

(40) Obligation:

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

IF2(true, x, y) → DIV(if(gt(x, y), x, y), y)
DIV(x0, 0) → IF1(true, x0, 0)
IF1(true, x, y) → IF2(gt(y, 0), x, y)
DIV(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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

(41) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF1(true, x, y) → IF2(gt(y, 0), x, y) at position [0] we obtained the following new rules [LPAR04]:

IF1(true, y0, 0) → IF2(false, y0, 0)
IF1(true, y0, s(x0)) → IF2(true, y0, s(x0))

(42) Obligation:

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

IF2(true, x, y) → DIV(if(gt(x, y), x, y), y)
DIV(x0, 0) → IF1(true, x0, 0)
DIV(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, y0, 0) → IF2(false, y0, 0)
IF1(true, y0, s(x0)) → IF2(true, y0, s(x0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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

(43) DependencyGraphProof (EQUIVALENT transformation)

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

(44) Obligation:

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

DIV(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, y0, s(x0)) → IF2(true, y0, s(x0))
IF2(true, x, y) → DIV(if(gt(x, y), x, y), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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

(45) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF2(true, x, y) → DIV(if(gt(x, y), x, y), y) at position [0] we obtained the following new rules [LPAR04]:

IF2(true, 0, x0) → DIV(if(false, 0, x0), x0)
IF2(true, s(x0), 0) → DIV(if(true, s(x0), 0), 0)
IF2(true, s(x0), s(x1)) → DIV(if(gt(x0, x1), s(x0), s(x1)), s(x1))

(46) Obligation:

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

DIV(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, y0, s(x0)) → IF2(true, y0, s(x0))
IF2(true, 0, x0) → DIV(if(false, 0, x0), x0)
IF2(true, s(x0), 0) → DIV(if(true, s(x0), 0), 0)
IF2(true, s(x0), s(x1)) → DIV(if(gt(x0, x1), s(x0), s(x1)), s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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

(47) DependencyGraphProof (EQUIVALENT transformation)

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

(48) Obligation:

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

IF1(true, y0, s(x0)) → IF2(true, y0, s(x0))
IF2(true, s(x0), s(x1)) → DIV(if(gt(x0, x1), s(x0), s(x1)), s(x1))
DIV(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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

(49) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF1(true, y0, s(x0)) → IF2(true, y0, s(x0)) we obtained the following new rules [LPAR04]:

IF1(true, s(z0), s(z1)) → IF2(true, s(z0), s(z1))

(50) Obligation:

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

IF2(true, s(x0), s(x1)) → DIV(if(gt(x0, x1), s(x0), s(x1)), s(x1))
DIV(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, s(z0), s(z1)) → IF2(true, s(z0), s(z1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
p(0)
p(s(x0))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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

(51) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(52) Obligation:

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

IF2(true, s(x0), s(x1)) → DIV(if(gt(x0, x1), s(x0), s(x1)), s(x1))
DIV(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, s(z0), s(z1)) → IF2(true, s(z0), s(z1))

The TRS R consists of the following rules:

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

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

(53) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(54) Obligation:

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

IF2(true, s(x0), s(x1)) → DIV(if(gt(x0, x1), s(x0), s(x1)), s(x1))
DIV(s(x0), s(x1)) → IF1(ge(x0, x1), s(x0), s(x1))
IF1(true, s(z0), s(z1)) → IF2(true, s(z0), s(z1))

The TRS R consists of the following rules:

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

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