YES Termination w.r.t. Q proof of Beerendonk_07_8.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 5 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 TransformationProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 DependencyGraphProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 TransformationProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 TransformationProof (⇔, 0 ms)
        ↳28 QDP
        ↳29 DependencyGraphProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 TransformationProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 TransformationProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 DependencyGraphProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 UsableRulesProof (⇔, 0 ms)
        ↳38 QDP
        ↳39 TransformationProof (⇔, 0 ms)
        ↳40 QDP
        ↳41 TransformationProof (⇔, 0 ms)
        ↳42 QDP
        ↳43 TransformationProof (⇔, 0 ms)
        ↳44 QDP
        ↳45 DependencyGraphProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 UsableRulesProof (⇔, 0 ms)
        ↳48 QDP
        ↳49 QReductionProof (⇔, 0 ms)
        ↳50 QDP
        ↳51 TransformationProof (⇔, 0 ms)
        ↳52 QDP
        ↳53 UsableRulesProof (⇔, 0 ms)
        ↳54 QDP
        ↳55 QReductionProof (⇔, 0 ms)
        ↳56 QDP
        ↳57 TransformationProof (⇔, 0 ms)
        ↳58 QDP
        ↳59 UsableRulesProof (⇔, 0 ms)
        ↳60 QDP
        ↳61 QReductionProof (⇔, 0 ms)
        ↳62 QDP
        ↳63 QDPSizeChangeProof (⇔, 0 ms)
        ↳64 YES

(0) Obligation:

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

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

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

The TRS R 2 is

cond(true, x, y) → cond(and(gr(x, 0), gr(y, 0)), p(x), p(y))

The signature Sigma is {cond}

(2) Obligation:

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

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

The set Q consists of the following terms:

cond(true, x0, x1)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(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:

COND(true, x, y) → COND(and(gr(x, 0), gr(y, 0)), p(x), p(y))
COND(true, x, y) → AND(gr(x, 0), gr(y, 0))
COND(true, x, y) → GR(x, 0)
COND(true, x, y) → GR(y, 0)
COND(true, x, y) → P(x)
COND(true, x, y) → P(y)
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond(true, x0, x1)
and(true, true)
and(x0, false)
and(false, x0)
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.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 5 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:

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

The set Q consists of the following terms:

cond(true, x0, x1)
and(true, true)
and(x0, false)
and(false, x0)
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.

(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:

cond(true, x0, x1)
and(true, true)
and(x0, false)
and(false, x0)
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.

(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].

cond(true, x0, x1)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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:

COND(true, x, y) → COND(and(gr(x, 0), gr(y, 0)), p(x), p(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond(true, x0, x1)
and(true, true)
and(x0, false)
and(false, x0)
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.

(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:

COND(true, x, y) → COND(and(gr(x, 0), gr(y, 0)), p(x), p(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond(true, x0, x1)
and(true, true)
and(x0, false)
and(false, x0)
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.

(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].

cond(true, x0, x1)

(18) Obligation:

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

COND(true, x, y) → COND(and(gr(x, 0), gr(y, 0)), p(x), p(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.

(19) TransformationProof (EQUIVALENT transformation)

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

COND(true, 0, y1) → COND(and(false, gr(y1, 0)), p(0), p(y1)) → COND(true, 0, y1) → COND(and(false, gr(y1, 0)), p(0), p(y1))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1)) → COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1))
COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0)) → COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0))) → COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))

(20) Obligation:

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

COND(true, 0, y1) → COND(and(false, gr(y1, 0)), p(0), p(y1))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1))
COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.

(21) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND(true, 0, y1) → COND(and(false, gr(y1, 0)), p(0), p(y1)) at position [0] we obtained the following new rules [LPAR04]:

COND(true, 0, y1) → COND(false, p(0), p(y1)) → COND(true, 0, y1) → COND(false, p(0), p(y1))

(22) Obligation:

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

COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1))
COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))
COND(true, 0, y1) → COND(false, p(0), p(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.

(23) DependencyGraphProof (EQUIVALENT transformation)

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

(24) Obligation:

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

COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1))
COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.

(25) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1)) at position [1] we obtained the following new rules [LPAR04]:

COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1)) → COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))

(26) Obligation:

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

COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.

(27) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0)) at position [0] we obtained the following new rules [LPAR04]:

COND(true, y0, 0) → COND(false, p(y0), p(0)) → COND(true, y0, 0) → COND(false, p(y0), p(0))

(28) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))
COND(true, y0, 0) → COND(false, p(y0), p(0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.

(29) DependencyGraphProof (EQUIVALENT transformation)

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

(30) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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 COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0))) at position [2] we obtained the following new rules [LPAR04]:

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0) → COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)

(32) Obligation:

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

COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1)) at position [0] we obtained the following new rules [LPAR04]:

COND(true, s(y0), 0) → COND(and(true, false), y0, p(0)) → COND(true, s(y0), 0) → COND(and(true, false), y0, p(0))
COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0))) → COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0)))

(34) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(y0), 0) → COND(and(true, false), y0, p(0))
COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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) DependencyGraphProof (EQUIVALENT transformation)

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

(36) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.

(37) 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.

(38) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.

(39) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0))) at position [0] we obtained the following new rules [LPAR04]:

COND(true, s(y0), s(x0)) → COND(true, y0, p(s(x0))) → COND(true, s(y0), s(x0)) → COND(true, y0, p(s(x0)))

(40) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(y0), s(x0)) → COND(true, y0, p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.

(41) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND(true, s(y0), s(x0)) → COND(true, y0, p(s(x0))) at position [2] we obtained the following new rules [LPAR04]:

COND(true, s(y0), s(x0)) → COND(true, y0, x0) → COND(true, s(y0), s(x0)) → COND(true, y0, x0)

(42) Obligation:

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(y0), s(x0)) → COND(true, y0, x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.

(43) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0) at position [0] we obtained the following new rules [LPAR04]:

COND(true, 0, s(y1)) → COND(and(false, true), p(0), y1) → COND(true, 0, s(y1)) → COND(and(false, true), p(0), y1)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1) → COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

(44) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, 0, s(y1)) → COND(and(false, true), p(0), y1)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.

(45) DependencyGraphProof (EQUIVALENT transformation)

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

(46) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.

(47) 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.

(48) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

The TRS R consists of the following rules:

and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
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.

(49) 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].

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

(50) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

The TRS R consists of the following rules:

and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
p(0)
p(s(x0))

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

(51) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1) at position [0] we obtained the following new rules [LPAR04]:

COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1) → COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1)

(52) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1)

The TRS R consists of the following rules:

and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
p(0)
p(s(x0))

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

(53) 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.

(54) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
p(0)
p(s(x0))

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

(55) 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].

and(true, true)
and(x0, false)
and(false, x0)

(56) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(57) TransformationProof (EQUIVALENT transformation)

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

COND(true, s(x0), s(y1)) → COND(true, x0, y1) → COND(true, s(y0), s(x0)) → COND(true, y0, x0)

(58) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

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

(59) 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.

(60) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)

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

p(0)
p(s(x0))

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

(61) 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))

(62) Obligation:

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)

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

(63) 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:

  • COND(true, s(y0), s(x0)) → COND(true, y0, x0)
    The graph contains the following edges 1 >= 1, 2 > 2, 3 > 3

(64) YES