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

0 QTRS
↳1 AAECC Innermost (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
    ↳19 Narrowing (⇔)
    ↳20 QDP
    ↳21 DependencyGraphProof (⇔)
    ↳22 QDP
    ↳23 Narrowing (⇔)
    ↳24 QDP
    ↳25 DependencyGraphProof (⇔)
    ↳26 QDP
    ↳27 UsableRulesProof (⇔)
    ↳28 QDP
    ↳29 QReductionProof (⇔)
    ↳30 QDP
    ↳31 Rewriting (⇔)
    ↳32 QDP
    ↳33 UsableRulesProof (⇔)
    ↳34 QDP
    ↳35 QReductionProof (⇔)
    ↳36 QDP
    ↳37 NonInfProof (⇔)
    ↳38 QDP
    ↳39 DependencyGraphProof (⇔)
    ↳40 TRUE

(0) Obligation:

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

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr0(x), y, y)
cond2(false, x, y) → cond1(gr0(x), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr0(0) → false
gr0(s(x)) → true
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)
gr0(0) → false
gr0(s(x)) → true
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr0(x), y, y)
cond2(false, x, y) → cond1(gr0(x), p(x), y)

The signature Sigma is {cond1, cond2}

(2) Obligation:

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

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr0(x), y, y)
cond2(false, x, y) → cond1(gr0(x), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr0(0) → false
gr0(s(x)) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
gr0(0)
gr0(s(x0))
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) → COND2(gr(x, y), x, y)
COND1(true, x, y) → GR(x, y)
COND2(true, x, y) → COND1(gr0(x), y, y)
COND2(true, x, y) → GR0(x)
COND2(false, x, y) → COND1(gr0(x), p(x), y)
COND2(false, x, y) → GR0(x)
COND2(false, x, y) → P(x)
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr0(x), y, y)
cond2(false, x, y) → cond1(gr0(x), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr0(0) → false
gr0(s(x)) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
gr0(0)
gr0(s(x0))
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 2 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) 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) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr0(x), y, y)
cond2(false, x, y) → cond1(gr0(x), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr0(0) → false
gr0(s(x)) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
gr0(0)
gr0(s(x0))
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:

GR(s(x), s(y)) → GR(x, y)

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

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
gr0(0)
gr0(s(x0))
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)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
gr0(0)
gr0(s(x0))
p(0)
p(s(x0))

(11) 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.

(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:

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

(13) TRUE

(14) Obligation:

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

COND2(true, x, y) → COND1(gr0(x), y, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND1(gr0(x), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(gr0(x), y, y)
cond2(false, x, y) → cond1(gr0(x), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr0(0) → false
gr0(s(x)) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
gr0(0)
gr0(s(x0))
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:

COND2(true, x, y) → COND1(gr0(x), y, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND1(gr0(x), p(x), y)

The TRS R consists of the following rules:

gr0(0) → false
gr0(s(x)) → true
p(0) → 0
p(s(x)) → x
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
gr0(0)
gr0(s(x0))
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)
cond2(true, x0, x1)
cond2(false, x0, x1)

(18) Obligation:

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

COND2(true, x, y) → COND1(gr0(x), y, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND1(gr0(x), p(x), y)

The TRS R consists of the following rules:

gr0(0) → false
gr0(s(x)) → true
p(0) → 0
p(s(x)) → x
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
gr0(0)
gr0(s(x0))
p(0)
p(s(x0))

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

(19) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND2(true, x, y) → COND1(gr0(x), y, y) at position [0] we obtained the following new rules [LPAR04]:

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

(20) Obligation:

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

COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND1(gr0(x), p(x), y)
COND2(true, 0, y1) → COND1(false, y1, y1)
COND2(true, s(x0), y1) → COND1(true, y1, y1)

The TRS R consists of the following rules:

gr0(0) → false
gr0(s(x)) → true
p(0) → 0
p(s(x)) → x
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
gr0(0)
gr0(s(x0))
p(0)
p(s(x0))

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) Obligation:

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

COND2(false, x, y) → COND1(gr0(x), p(x), y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(true, s(x0), y1) → COND1(true, y1, y1)

The TRS R consists of the following rules:

gr0(0) → false
gr0(s(x)) → true
p(0) → 0
p(s(x)) → x
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
gr0(0)
gr0(s(x0))
p(0)
p(s(x0))

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

(23) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND2(false, x, y) → COND1(gr0(x), p(x), y) at position [0] we obtained the following new rules [LPAR04]:

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

(24) Obligation:

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

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

The TRS R consists of the following rules:

gr0(0) → false
gr0(s(x)) → true
p(0) → 0
p(s(x)) → x
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
gr0(0)
gr0(s(x0))
p(0)
p(s(x0))

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

(25) DependencyGraphProof (EQUIVALENT transformation)

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

(26) Obligation:

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

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

The TRS R consists of the following rules:

gr0(0) → false
gr0(s(x)) → true
p(0) → 0
p(s(x)) → x
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
gr0(0)
gr0(s(x0))
p(0)
p(s(x0))

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

(27) 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.

(28) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
gr0(0)
gr0(s(x0))
p(0)
p(s(x0))

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

(29) 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].

gr0(0)
gr0(s(x0))

(30) Obligation:

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

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

The TRS R consists of the following rules:

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

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.

(31) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND2(false, s(x0), y1) → COND1(true, p(s(x0)), y1) at position [1] we obtained the following new rules [LPAR04]:

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

(32) Obligation:

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

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

The TRS R consists of the following rules:

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

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.

(33) 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.

(34) Obligation:

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

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

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)

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.

(35) 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))

(36) Obligation:

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

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

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)

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.

(37) 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 COND2(true, s(x0), y1) → COND1(true, y1, y1) the following chains were created:
  • We consider the chain COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(true, s(x0), y1) → COND1(true, y1, y1) which results in the following constraint:

    (1)    (COND2(gr(x2, x3), x2, x3)=COND2(true, s(x4), x5) ⇒ COND2(true, s(x4), x5)≥COND1(true, x5, x5))



    We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (2)    (s(x4)=x26∧gr(x26, x3)=true ⇒ COND2(true, s(x4), x3)≥COND1(true, x3, x3))



    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x26, x3)=true which results in the following new constraints:

    (3)    (true=true∧s(x4)=s(x28) ⇒ COND2(true, s(x4), 0)≥COND1(true, 0, 0))


    (4)    (gr(x30, x29)=true∧s(x4)=s(x30)∧(∀x31:gr(x30, x29)=true∧s(x31)=x30 ⇒ COND2(true, s(x31), x29)≥COND1(true, x29, x29)) ⇒ COND2(true, s(x4), s(x29))≥COND1(true, s(x29), s(x29)))



    We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint:

    (5)    (COND2(true, s(x4), 0)≥COND1(true, 0, 0))



    We simplified constraint (4) using rules (I), (II), (III), (IV) which results in the following new constraint:

    (6)    (gr(x30, x29)=true ⇒ COND2(true, s(x30), s(x29))≥COND1(true, s(x29), s(x29)))



    We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on gr(x30, x29)=true which results in the following new constraints:

    (7)    (true=true ⇒ COND2(true, s(s(x33)), s(0))≥COND1(true, s(0), s(0)))


    (8)    (gr(x35, x34)=true∧(gr(x35, x34)=true ⇒ COND2(true, s(x35), s(x34))≥COND1(true, s(x34), s(x34))) ⇒ COND2(true, s(s(x35)), s(s(x34)))≥COND1(true, s(s(x34)), s(s(x34))))



    We simplified constraint (7) using rules (I), (II) which results in the following new constraint:

    (9)    (COND2(true, s(s(x33)), s(0))≥COND1(true, s(0), s(0)))



    We simplified constraint (8) using rule (VI) where we applied the induction hypothesis (gr(x35, x34)=true ⇒ COND2(true, s(x35), s(x34))≥COND1(true, s(x34), s(x34))) with σ = [ ] which results in the following new constraint:

    (10)    (COND2(true, s(x35), s(x34))≥COND1(true, s(x34), s(x34)) ⇒ COND2(true, s(s(x35)), s(s(x34)))≥COND1(true, s(s(x34)), s(s(x34))))







For Pair COND1(true, x, y) → COND2(gr(x, y), x, y) the following chains were created:
  • We consider the chain COND2(true, s(x0), y1) → COND1(true, y1, y1), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:

    (11)    (COND1(true, x9, x9)=COND1(true, x10, x11) ⇒ COND1(true, x10, x11)≥COND2(gr(x10, x11), x10, x11))



    We simplified constraint (11) using rules (I), (II), (III) which results in the following new constraint:

    (12)    (COND1(true, x9, x9)≥COND2(gr(x9, x9), x9, x9))



  • We consider the chain COND2(false, s(x0), y1) → COND1(true, x0, y1), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:

    (13)    (COND1(true, x14, x15)=COND1(true, x16, x17) ⇒ COND1(true, x16, x17)≥COND2(gr(x16, x17), x16, x17))



    We simplified constraint (13) using rules (I), (II), (III) which results in the following new constraint:

    (14)    (COND1(true, x14, x15)≥COND2(gr(x14, x15), x14, x15))







For Pair COND2(false, s(x0), y1) → COND1(true, x0, y1) the following chains were created:
  • We consider the chain COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(false, s(x0), y1) → COND1(true, x0, y1) which results in the following constraint:

    (15)    (COND2(gr(x20, x21), x20, x21)=COND2(false, s(x22), x23) ⇒ COND2(false, s(x22), x23)≥COND1(true, x22, x23))



    We simplified constraint (15) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (16)    (s(x22)=x36∧gr(x36, x21)=false ⇒ COND2(false, s(x22), x21)≥COND1(true, x22, x21))



    We simplified constraint (16) using rule (V) (with possible (I) afterwards) using induction on gr(x36, x21)=false which results in the following new constraints:

    (17)    (false=false∧s(x22)=0 ⇒ COND2(false, s(x22), x37)≥COND1(true, x22, x37))


    (18)    (gr(x40, x39)=false∧s(x22)=s(x40)∧(∀x41:gr(x40, x39)=false∧s(x41)=x40 ⇒ COND2(false, s(x41), x39)≥COND1(true, x41, x39)) ⇒ COND2(false, s(x22), s(x39))≥COND1(true, x22, s(x39)))



    We solved constraint (17) using rules (I), (II).We simplified constraint (18) using rules (I), (II), (III), (IV) which results in the following new constraint:

    (19)    (gr(x40, x39)=false ⇒ COND2(false, s(x40), s(x39))≥COND1(true, x40, s(x39)))



    We simplified constraint (19) using rule (V) (with possible (I) afterwards) using induction on gr(x40, x39)=false which results in the following new constraints:

    (20)    (false=false ⇒ COND2(false, s(0), s(x42))≥COND1(true, 0, s(x42)))


    (21)    (gr(x45, x44)=false∧(gr(x45, x44)=false ⇒ COND2(false, s(x45), s(x44))≥COND1(true, x45, s(x44))) ⇒ COND2(false, s(s(x45)), s(s(x44)))≥COND1(true, s(x45), s(s(x44))))



    We simplified constraint (20) using rules (I), (II) which results in the following new constraint:

    (22)    (COND2(false, s(0), s(x42))≥COND1(true, 0, s(x42)))



    We simplified constraint (21) using rule (VI) where we applied the induction hypothesis (gr(x45, x44)=false ⇒ COND2(false, s(x45), s(x44))≥COND1(true, x45, s(x44))) with σ = [ ] which results in the following new constraint:

    (23)    (COND2(false, s(x45), s(x44))≥COND1(true, x45, s(x44)) ⇒ COND2(false, s(s(x45)), s(s(x44)))≥COND1(true, s(x45), s(s(x44))))







To summarize, we get the following constraints P≥ for the following pairs.
  • COND2(true, s(x0), y1) → COND1(true, y1, y1)
    • (COND2(true, s(x4), 0)≥COND1(true, 0, 0))
    • (COND2(true, s(s(x33)), s(0))≥COND1(true, s(0), s(0)))
    • (COND2(true, s(x35), s(x34))≥COND1(true, s(x34), s(x34)) ⇒ COND2(true, s(s(x35)), s(s(x34)))≥COND1(true, s(s(x34)), s(s(x34))))

  • COND1(true, x, y) → COND2(gr(x, y), x, y)
    • (COND1(true, x9, x9)≥COND2(gr(x9, x9), x9, x9))
    • (COND1(true, x14, x15)≥COND2(gr(x14, x15), x14, x15))

  • COND2(false, s(x0), y1) → COND1(true, x0, y1)
    • (COND2(false, s(0), s(x42))≥COND1(true, 0, s(x42)))
    • (COND2(false, s(x45), s(x44))≥COND1(true, x45, s(x44)) ⇒ COND2(false, s(s(x45)), s(s(x44)))≥COND1(true, s(x45), s(s(x44))))




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

The following pairs are in P>:

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

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

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

(38) Obligation:

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

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

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)

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.

(39) DependencyGraphProof (EQUIVALENT transformation)

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

(40) TRUE