YES Termination w.r.t. Q proof of AProVE_08_log.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPSizeChangeProof (⇔, 0 ms)
        ↳9 YES
    ↳10 QDP
        ↳11 NonMonReductionPairProof (⇔, 96 ms)
        ↳12 QDP
        ↳13 PisEmptyProof (⇔, 0 ms)
        ↳14 YES
    ↳15 QDP
        ↳16 NonMonReductionPairProof (⇔, 70 ms)
        ↳17 QDP
        ↳18 PisEmptyProof (⇔, 0 ms)
        ↳19 YES

(0) Obligation:

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

p(0) → 0
p(s(x)) → x
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
minus(x, s(y)) → p(minus(x, y))
div(0, s(y)) → 0
div(s(x), s(y)) → s(div(minus(s(x), s(y)), s(y)))
log(s(0), s(s(y))) → 0
log(s(s(x)), s(s(y))) → s(log(div(minus(x, y), s(s(y))), s(s(y))))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)
MINUS(x, s(y)) → P(minus(x, y))
MINUS(x, s(y)) → MINUS(x, y)
DIV(s(x), s(y)) → DIV(minus(s(x), s(y)), s(y))
DIV(s(x), s(y)) → MINUS(s(x), s(y))
LOG(s(s(x)), s(s(y))) → LOG(div(minus(x, y), s(s(y))), s(s(y)))
LOG(s(s(x)), s(s(y))) → DIV(minus(x, y), s(s(y)))
LOG(s(s(x)), s(s(y))) → MINUS(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
minus(x, s(y)) → p(minus(x, y))
div(0, s(y)) → 0
div(s(x), s(y)) → s(div(minus(s(x), s(y)), s(y)))
log(s(0), s(s(y))) → 0
log(s(s(x)), s(s(y))) → s(log(div(minus(x, y), s(s(y))), s(s(y))))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 4 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

MINUS(x, s(y)) → MINUS(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
minus(x, s(y)) → p(minus(x, y))
div(0, s(y)) → 0
div(s(x), s(y)) → s(div(minus(s(x), s(y)), s(y)))
log(s(0), s(s(y))) → 0
log(s(s(x)), s(s(y))) → s(log(div(minus(x, y), s(s(y))), s(s(y))))

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

MINUS(x, s(y)) → MINUS(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)

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

(8) 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:

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

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

(9) YES

(10) Obligation:

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

DIV(s(x), s(y)) → DIV(minus(s(x), s(y)), s(y))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
minus(x, s(y)) → p(minus(x, y))
div(0, s(y)) → 0
div(s(x), s(y)) → s(div(minus(s(x), s(y)), s(y)))
log(s(0), s(s(y))) → 0
log(s(s(x)), s(s(y))) → s(log(div(minus(x, y), s(s(y))), s(s(y))))

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

(11) NonMonReductionPairProof (EQUIVALENT transformation)

Using the following max-polynomial ordering, we can orient the general usable rules and all rules from P weakly and some rules from P strictly:
Polynomial interpretation with max [POLO,NEGPOLO,MAXPOLO]:

POL(0) = 0   
POL(DIV(x1, x2)) = x1   
POL(div(x1, x2)) = 0   
POL(log(x1, x2)) = 0   
POL(minus(x1, x2)) = max(0, x1 - x2)   
POL(p(x1)) = max(0, -1 + x1)   
POL(s(x1)) = 1 + x1   

The following pairs can be oriented strictly and are deleted.

DIV(s(x), s(y)) → DIV(minus(s(x), s(y)), s(y))
The remaining pairs can at least be oriented weakly.
none
The following rules are usable:

p(0) → 0
p(s(x)) → x
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
minus(x, s(y)) → p(minus(x, y))

(12) Obligation:

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

p(0) → 0
p(s(x)) → x
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
minus(x, s(y)) → p(minus(x, y))
div(0, s(y)) → 0
div(s(x), s(y)) → s(div(minus(s(x), s(y)), s(y)))
log(s(0), s(s(y))) → 0
log(s(s(x)), s(s(y))) → s(log(div(minus(x, y), s(s(y))), s(s(y))))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) YES

(15) Obligation:

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

LOG(s(s(x)), s(s(y))) → LOG(div(minus(x, y), s(s(y))), s(s(y)))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
minus(x, s(y)) → p(minus(x, y))
div(0, s(y)) → 0
div(s(x), s(y)) → s(div(minus(s(x), s(y)), s(y)))
log(s(0), s(s(y))) → 0
log(s(s(x)), s(s(y))) → s(log(div(minus(x, y), s(s(y))), s(s(y))))

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

(16) NonMonReductionPairProof (EQUIVALENT transformation)

Using the following max-polynomial ordering, we can orient the general usable rules and all rules from P weakly and some rules from P strictly:
Polynomial interpretation with max [POLO,NEGPOLO,MAXPOLO]:

POL(0) = 0   
POL(LOG(x1, x2)) = x1   
POL(div(x1, x2)) = x1   
POL(log(x1, x2)) = 1 + x1 + x2   
POL(minus(x1, x2)) = max(0, x1 - x2)   
POL(p(x1)) = max(0, -1 + x1)   
POL(s(x1)) = 1 + x1   

The following pairs can be oriented strictly and are deleted.

LOG(s(s(x)), s(s(y))) → LOG(div(minus(x, y), s(s(y))), s(s(y)))
The remaining pairs can at least be oriented weakly.
none
The following rules are usable:

p(0) → 0
p(s(x)) → x
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
minus(x, s(y)) → p(minus(x, y))
div(0, s(y)) → 0
div(s(x), s(y)) → s(div(minus(s(x), s(y)), s(y)))

(17) Obligation:

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

p(0) → 0
p(s(x)) → x
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
minus(x, s(y)) → p(minus(x, y))
div(0, s(y)) → 0
div(s(x), s(y)) → s(div(minus(s(x), s(y)), s(y)))
log(s(0), s(s(y))) → 0
log(s(s(x)), s(s(y))) → s(log(div(minus(x, y), s(s(y))), s(s(y))))

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) YES