Termination w.r.t. Q of the given QTRS could not be shown:

(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x2)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

s₁ > MINUS₁

The following usable rules [FROCOS05] were oriented: none

(7) 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.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(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) 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.