YES Termination w.r.t. Q proof of Beerendonk_07_14.ari

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 DependencyGraphProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 UsableRulesProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 QReductionProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 NonInfProof (⇔, 20 ms)
        ↳35 QDP
        ↳36 DependencyGraphProof (⇔, 0 ms)
        ↳37 TRUE

(0) Obligation:

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

cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(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)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)

The signature Sigma is {cond1, cond2}

(2) Obligation:

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

cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(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(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND1(s(x), y) → GR(s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, x, y) → COND1(p(x), y)
COND2(false, x, y) → P(x)
GR(s(x), s(y)) → GR(x, y)
NEQ(s(x), s(y)) → NEQ(x, y)

The TRS R consists of the following rules:

cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(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 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

NEQ(s(x), s(y)) → NEQ(x, y)

The TRS R consists of the following rules:

cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(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:

NEQ(s(x), s(y)) → NEQ(x, y)

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

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(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(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

(11) Obligation:

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

NEQ(s(x), s(y)) → NEQ(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:

  • NEQ(s(x), s(y)) → NEQ(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(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(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(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(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(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(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) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, x, y) → COND1(p(x), y)

The TRS R consists of the following rules:

cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(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) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, x, y) → COND1(p(x), y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
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:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(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(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))

(25) Obligation:

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

COND2(true, x, y) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, x, y) → COND1(p(x), y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
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.

(26) TransformationProof (EQUIVALENT transformation)

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

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

(27) Obligation:

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

COND2(true, x, y) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, 0, y1) → COND1(0, y1)
COND2(false, s(x0), y1) → COND1(x0, y1)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
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.

(28) DependencyGraphProof (EQUIVALENT transformation)

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

(29) Obligation:

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

COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
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.

(30) 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.

(31) Obligation:

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

COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)

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.

(32) 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))

(33) Obligation:

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

COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)

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.

(34) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair COND1(s(x), y) → COND2(gr(s(x), y), s(x), y) the following chains were created:
  • We consider the chain COND2(true, x2, x3) → COND1(x3, x3), COND1(s(x4), x5) → COND2(gr(s(x4), x5), s(x4), x5) which results in the following constraint:
    (1)    (COND1(x3, x3)=COND1(s(x4), x5) ⇒ COND1(s(x4), x5)≥COND2(gr(s(x4), x5), s(x4), x5))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (COND1(s(x4), s(x4))≥COND2(gr(s(x4), s(x4)), s(x4), s(x4)))


  • We consider the chain COND2(false, s(x6), x7) → COND1(x6, x7), COND1(s(x8), x9) → COND2(gr(s(x8), x9), s(x8), x9) which results in the following constraint:
    (1)    (COND1(x6, x7)=COND1(s(x8), x9) ⇒ COND1(s(x8), x9)≥COND2(gr(s(x8), x9), s(x8), x9))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (COND1(s(x8), x7)≥COND2(gr(s(x8), x7), s(x8), x7))






For Pair COND2(true, x, y) → COND1(y, y) the following chains were created:
  • We consider the chain COND1(s(x10), x11) → COND2(gr(s(x10), x11), s(x10), x11), COND2(true, x12, x13) → COND1(x13, x13) which results in the following constraint:
    (1)    (COND2(gr(s(x10), x11), s(x10), x11)=COND2(true, x12, x13) ⇒ COND2(true, x12, x13)≥COND1(x13, x13))


    We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
    (2)    (s(x10)=x26∧gr(x26, x11)=true ⇒ COND2(true, s(x10), x11)≥COND1(x11, x11))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x26, x11)=true which results in the following new constraints:
    (3)    (true=true∧s(x10)=s(x27) ⇒ COND2(true, s(x10), 0)≥COND1(0, 0))

    (4)    (gr(x29, x28)=true∧s(x10)=s(x29)∧(∀x30:gr(x29, x28)=true∧s(x30)=x29 ⇒ COND2(true, s(x30), x28)≥COND1(x28, x28)) ⇒ COND2(true, s(x10), s(x28))≥COND1(s(x28), s(x28)))


    We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint:
    (5)    (COND2(true, s(x10), 0)≥COND1(0, 0))


    We simplified constraint (4) using rules (I), (II), (III), (IV) which results in the following new constraint:
    (6)    (gr(x29, x28)=true ⇒ COND2(true, s(x29), s(x28))≥COND1(s(x28), s(x28)))


    We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on gr(x29, x28)=true which results in the following new constraints:
    (7)    (true=true ⇒ COND2(true, s(s(x32)), s(0))≥COND1(s(0), s(0)))

    (8)    (gr(x34, x33)=true∧(gr(x34, x33)=true ⇒ COND2(true, s(x34), s(x33))≥COND1(s(x33), s(x33))) ⇒ COND2(true, s(s(x34)), s(s(x33)))≥COND1(s(s(x33)), s(s(x33))))


    We simplified constraint (7) using rules (I), (II) which results in the following new constraint:
    (9)    (COND2(true, s(s(x32)), s(0))≥COND1(s(0), s(0)))


    We simplified constraint (8) using rule (VI) where we applied the induction hypothesis (gr(x34, x33)=true ⇒ COND2(true, s(x34), s(x33))≥COND1(s(x33), s(x33))) with σ = [ ] which results in the following new constraint:
    (10)    (COND2(true, s(x34), s(x33))≥COND1(s(x33), s(x33)) ⇒ COND2(true, s(s(x34)), s(s(x33)))≥COND1(s(s(x33)), s(s(x33))))






For Pair COND2(false, s(x0), y1) → COND1(x0, y1) the following chains were created:
  • We consider the chain COND1(s(x18), x19) → COND2(gr(s(x18), x19), s(x18), x19), COND2(false, s(x20), x21) → COND1(x20, x21) which results in the following constraint:
    (1)    (COND2(gr(s(x18), x19), s(x18), x19)=COND2(false, s(x20), x21) ⇒ COND2(false, s(x20), x21)≥COND1(x20, x21))


    We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
    (2)    (s(x18)=x36∧gr(x36, x19)=false ⇒ COND2(false, s(x18), x19)≥COND1(x18, x19))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x36, x19)=false which results in the following new constraints:
    (3)    (gr(x39, x38)=false∧s(x18)=s(x39)∧(∀x40:gr(x39, x38)=false∧s(x40)=x39 ⇒ COND2(false, s(x40), x38)≥COND1(x40, x38)) ⇒ COND2(false, s(x18), s(x38))≥COND1(x18, s(x38)))

    (4)    (false=false∧s(x18)=0 ⇒ COND2(false, s(x18), x41)≥COND1(x18, x41))


    We simplified constraint (3) using rules (I), (II), (III), (IV) which results in the following new constraint:
    (5)    (gr(x39, x38)=false ⇒ COND2(false, s(x39), s(x38))≥COND1(x39, s(x38)))


    We solved constraint (4) using rules (I), (II).We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on gr(x39, x38)=false which results in the following new constraints:
    (6)    (gr(x44, x43)=false∧(gr(x44, x43)=false ⇒ COND2(false, s(x44), s(x43))≥COND1(x44, s(x43))) ⇒ COND2(false, s(s(x44)), s(s(x43)))≥COND1(s(x44), s(s(x43))))

    (7)    (false=false ⇒ COND2(false, s(0), s(x45))≥COND1(0, s(x45)))


    We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (gr(x44, x43)=false ⇒ COND2(false, s(x44), s(x43))≥COND1(x44, s(x43))) with σ = [ ] which results in the following new constraint:
    (8)    (COND2(false, s(x44), s(x43))≥COND1(x44, s(x43)) ⇒ COND2(false, s(s(x44)), s(s(x43)))≥COND1(s(x44), s(s(x43))))


    We simplified constraint (7) using rules (I), (II) which results in the following new constraint:
    (9)    (COND2(false, s(0), s(x45))≥COND1(0, s(x45)))






To summarize, we get the following constraints P≥ for the following pairs.
  • COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
    • (COND1(s(x4), s(x4))≥COND2(gr(s(x4), s(x4)), s(x4), s(x4)))
    • (COND1(s(x8), x7)≥COND2(gr(s(x8), x7), s(x8), x7))

  • COND2(true, x, y) → COND1(y, y)
    • (COND2(true, s(x10), 0)≥COND1(0, 0))
    • (COND2(true, s(s(x32)), s(0))≥COND1(s(0), s(0)))
    • (COND2(true, s(x34), s(x33))≥COND1(s(x33), s(x33)) ⇒ COND2(true, s(s(x34)), s(s(x33)))≥COND1(s(s(x33)), s(s(x33))))

  • COND2(false, s(x0), y1) → COND1(x0, y1)
    • (COND2(false, s(x44), s(x43))≥COND1(x44, s(x43)) ⇒ COND2(false, s(s(x44)), s(s(x43)))≥COND1(s(x44), s(s(x43))))
    • (COND2(false, s(0), s(x45))≥COND1(0, s(x45)))




The constraints for P> respective Pbound are constructed from P≥ where we just replace every occurence of "t ≥ s" in P≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(COND1(x1, x2)) = x1   
POL(COND2(x1, x2, x3)) = -1 + x1 + x2   
POL(c) = -1   
POL(false) = 1   
POL(gr(x1, x2)) = 1   
POL(s(x1)) = 1 + x1   
POL(true) = 1   

The following pairs are in P>:

COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)
The following pairs are in Pbound:

COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)
The following rules are usable:

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

(35) Obligation:

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

COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)

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.

(36) DependencyGraphProof (EQUIVALENT transformation)

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

(37) TRUE