YES Termination w.r.t. Q proof of Beerendonk_07_12.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 10 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 TransformationProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 DependencyGraphProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 TransformationProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 DependencyGraphProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 UsableRulesProof (⇔, 0 ms)
        ↳28 QDP
        ↳29 QReductionProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 TransformationProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 UsableRulesProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QReductionProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 NonInfProof (⇔, 0 ms)
        ↳38 AND
            ↳39 QDP
                ↳40 TransformationProof (⇔, 0 ms)
                ↳41 QDP
                ↳42 TransformationProof (⇔, 0 ms)
                ↳43 QDP
                ↳44 QDPQMonotonicMRRProof (⇔, 1 ms)
                ↳45 QDP
                ↳46 NonInfProof (⇔, 0 ms)
                ↳47 QDP
                ↳48 PisEmptyProof (⇔, 0 ms)
                ↳49 YES
            ↳50 QDP
                ↳51 QDPSizeChangeProof (⇔, 0 ms)
                ↳52 YES

(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) YES

(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) TransformationProof (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, 0, y1) → COND1(false, y1, y1)
COND2(true, s(x0), y1) → COND1(true, 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) TransformationProof (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, 0, y1) → COND1(false, p(0), y1)
COND2(false, s(x0), y1) → COND1(true, p(s(x0)), 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) TransformationProof (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) → 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, x2, x3) → COND2(gr(x2, x3), x2, x3), COND2(true, s(x4), x5) → COND1(true, x5, x5) 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(x8), x9) → COND1(true, x9, x9), COND1(true, x10, x11) → COND2(gr(x10, x11), x10, x11) which results in the following constraint:
    (1)    (COND1(true, x9, x9)=COND1(true, x10, x11) ⇒ COND1(true, x10, x11)≥COND2(gr(x10, x11), x10, x11))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (COND1(true, x9, x9)≥COND2(gr(x9, x9), x9, x9))


  • We consider the chain COND2(false, s(x14), x15) → COND1(true, x14, x15), COND1(true, x16, x17) → COND2(gr(x16, x17), x16, x17) which results in the following constraint:
    (1)    (COND1(true, x14, x15)=COND1(true, x16, x17) ⇒ COND1(true, x16, x17)≥COND2(gr(x16, x17), x16, x17))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (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, x20, x21) → COND2(gr(x20, x21), x20, x21), COND2(false, s(x22), x23) → COND1(true, x22, x23) which results in the following constraint:
    (1)    (COND2(gr(x20, x21), x20, x21)=COND2(false, s(x22), x23) ⇒ COND2(false, s(x22), x23)≥COND1(true, x22, x23))


    We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
    (2)    (s(x22)=x36∧gr(x36, x21)=false ⇒ COND2(false, s(x22), x21)≥COND1(true, x22, x21))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x36, x21)=false which results in the following new constraints:
    (3)    (false=false∧s(x22)=0 ⇒ COND2(false, s(x22), x37)≥COND1(true, x22, x37))

    (4)    (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 (3) using rules (I), (II).We simplified constraint (4) using rules (I), (II), (III), (IV) which results in the following new constraint:
    (5)    (gr(x40, x39)=false ⇒ COND2(false, s(x40), s(x39))≥COND1(true, x40, s(x39)))


    We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on gr(x40, x39)=false which results in the following new constraints:
    (6)    (false=false ⇒ COND2(false, s(0), s(x42))≥COND1(true, 0, s(x42)))

    (7)    (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 (6) using rules (I), (II) which results in the following new constraint:
    (8)    (COND2(false, s(0), s(x42))≥COND1(true, 0, s(x42)))


    We simplified constraint (7) 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:
    (9)    (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)) = -x1 + x2 - x3   
POL(COND2(x1, x2, x3)) = -1 + x1 + x2 - x3   
POL(c) = -1   
POL(false) = 1   
POL(gr(x1, x2)) = 1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following pairs are in P>:

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

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

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

(38) Complex Obligation (AND)

(39) 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)

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.

(40) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND1(true, x, y) → COND2(gr(x, y), x, y) we obtained the following new rules [LPAR04]:

COND1(true, z1, z1) → COND2(gr(z1, z1), z1, z1) → COND1(true, z1, z1) → COND2(gr(z1, z1), z1, z1)

(41) Obligation:

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

COND2(true, s(x0), y1) → COND1(true, y1, y1)
COND1(true, z1, z1) → COND2(gr(z1, z1), z1, 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)

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.

(42) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND2(true, s(x0), y1) → COND1(true, y1, y1) we obtained the following new rules [LPAR04]:

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

(43) Obligation:

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

COND1(true, z1, z1) → COND2(gr(z1, z1), z1, z1)
COND2(true, s(x0), s(x0)) → COND1(true, s(x0), s(x0))

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.

(44) QDPQMonotonicMRRProof (EQUIVALENT transformation)

By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain.

Strictly oriented rules of the TRS R:

gr(0, x) → false

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(COND1(x1, x2, x3)) = 2·x1 + x2 + 2·x3   
POL(COND2(x1, x2, x3)) = 2 + x1 + x2 + 2·x3   
POL(false) = 0   
POL(gr(x1, x2)) = 2   
POL(s(x1)) = x1   
POL(true) = 2   

(45) Obligation:

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

COND1(true, z1, z1) → COND2(gr(z1, z1), z1, z1)
COND2(true, s(x0), s(x0)) → COND1(true, s(x0), s(x0))

The TRS R consists of the following rules:

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.

(46) 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(true, z1, z1) → COND2(gr(z1, z1), z1, z1) the following chains were created:
  • We consider the chain COND2(true, s(x1), s(x1)) → COND1(true, s(x1), s(x1)), COND1(true, x2, x2) → COND2(gr(x2, x2), x2, x2) which results in the following constraint:
    (1)    (COND1(true, s(x1), s(x1))=COND1(true, x2, x2) ⇒ COND1(true, x2, x2)≥COND2(gr(x2, x2), x2, x2))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (COND1(true, s(x1), s(x1))≥COND2(gr(s(x1), s(x1)), s(x1), s(x1)))






For Pair COND2(true, s(x0), s(x0)) → COND1(true, s(x0), s(x0)) the following chains were created:
  • We consider the chain COND1(true, x3, x3) → COND2(gr(x3, x3), x3, x3), COND2(true, s(x4), s(x4)) → COND1(true, s(x4), s(x4)) which results in the following constraint:
    (1)    (COND2(gr(x3, x3), x3, x3)=COND2(true, s(x4), s(x4)) ⇒ COND2(true, s(x4), s(x4))≥COND1(true, s(x4), s(x4)))


    We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
    (2)    (s(x4)=x6∧s(x4)=x7∧gr(x6, x7)=true ⇒ COND2(true, s(x4), s(x4))≥COND1(true, s(x4), s(x4)))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x6, x7)=true which results in the following new constraints:
    (3)    (true=true∧s(x4)=s(x8)∧s(x4)=0 ⇒ COND2(true, s(x4), s(x4))≥COND1(true, s(x4), s(x4)))

    (4)    (gr(x10, x9)=true∧s(x4)=s(x10)∧s(x4)=s(x9)∧(∀x11:gr(x10, x9)=true∧s(x11)=x10∧s(x11)=x9 ⇒ COND2(true, s(x11), s(x11))≥COND1(true, s(x11), s(x11))) ⇒ COND2(true, s(x4), s(x4))≥COND1(true, s(x4), s(x4)))


    We solved constraint (3) using rules (I), (II).We simplified constraint (4) using rules (I), (II), (III), (IV) which results in the following new constraint:
    (5)    (gr(x10, x9)=true∧x10=x9 ⇒ COND2(true, s(x10), s(x10))≥COND1(true, s(x10), s(x10)))


    We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on gr(x10, x9)=true which results in the following new constraints:
    (6)    (true=true∧s(x12)=0 ⇒ COND2(true, s(s(x12)), s(s(x12)))≥COND1(true, s(s(x12)), s(s(x12))))

    (7)    (gr(x14, x13)=true∧s(x14)=s(x13)∧(gr(x14, x13)=true∧x14=x13 ⇒ COND2(true, s(x14), s(x14))≥COND1(true, s(x14), s(x14))) ⇒ COND2(true, s(s(x14)), s(s(x14)))≥COND1(true, s(s(x14)), s(s(x14))))


    We solved constraint (6) using rules (I), (II).We simplified constraint (7) using rules (I), (II) which results in the following new constraint:
    (8)    (gr(x14, x13)=true∧x14=x13∧(gr(x14, x13)=true∧x14=x13 ⇒ COND2(true, s(x14), s(x14))≥COND1(true, s(x14), s(x14))) ⇒ COND2(true, s(s(x14)), s(s(x14)))≥COND1(true, s(s(x14)), s(s(x14))))


    We simplified constraint (8) using rule (VI) where we applied the induction hypothesis (gr(x14, x13)=true∧x14=x13 ⇒ COND2(true, s(x14), s(x14))≥COND1(true, s(x14), s(x14))) with σ = [ ] which results in the following new constraint:
    (9)    (COND2(true, s(x14), s(x14))≥COND1(true, s(x14), s(x14)) ⇒ COND2(true, s(s(x14)), s(s(x14)))≥COND1(true, s(s(x14)), s(s(x14))))






To summarize, we get the following constraints P≥ for the following pairs.
  • COND1(true, z1, z1) → COND2(gr(z1, z1), z1, z1)
    • (COND1(true, s(x1), s(x1))≥COND2(gr(s(x1), s(x1)), s(x1), s(x1)))

  • COND2(true, s(x0), s(x0)) → COND1(true, s(x0), s(x0))
    • (COND2(true, s(x14), s(x14))≥COND1(true, s(x14), s(x14)) ⇒ COND2(true, s(s(x14)), s(s(x14)))≥COND1(true, s(s(x14)), s(s(x14))))




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

The following pairs are in P>:

COND1(true, z1, z1) → COND2(gr(z1, z1), z1, z1)
COND2(true, s(x0), s(x0)) → COND1(true, s(x0), s(x0))
The following pairs are in Pbound:

COND1(true, z1, z1) → COND2(gr(z1, z1), z1, z1)
COND2(true, s(x0), s(x0)) → COND1(true, s(x0), s(x0))
The following rules are usable:

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

(47) Obligation:

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

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.

(48) PisEmptyProof (EQUIVALENT transformation)

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

(49) YES

(50) Obligation:

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

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.

(51) 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, s(x0), y1) → COND1(true, x0, y1)
    The graph contains the following edges 2 > 2, 3 >= 3

  • COND1(true, x, y) → COND2(gr(x, y), x, y)
    The graph contains the following edges 2 >= 2, 3 >= 3

(52) YES