Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/variousSLASH15.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f₁(s₁(x)) -> s₁(s₁(f₁(p₁(s₁(x)))))
f₁(0) -> 0
p₁(s₁(x)) -> x

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

f₁(s₁(x)) -> s₁(s₁(f₁(p₁(s₁(x)))))
f₁(0) -> 0
p₁(s₁(x)) -> x

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

f₁(s₁(x)) -> s₁(s₁(f₁(p₁(s₁(x)))))
f₁(0) -> 0
p₁(s₁(x)) -> x

The set Q consists of the following terms:

f₁(s₁(x0))
f₁(0)
p₁(s₁(x0))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F₁(s₁(x)) -> P₁(s₁(x))
F₁(s₁(x)) -> F₁(p₁(s₁(x)))

The TRS R consists of the following rules:

f₁(s₁(x)) -> s₁(s₁(f₁(p₁(s₁(x)))))
f₁(0) -> 0
p₁(s₁(x)) -> x

The set Q consists of the following terms:

f₁(s₁(x0))
f₁(0)
p₁(s₁(x0))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

F₁(s₁(x)) -> P₁(s₁(x))
F₁(s₁(x)) -> F₁(p₁(s₁(x)))

The TRS R consists of the following rules:

f₁(s₁(x)) -> s₁(s₁(f₁(p₁(s₁(x)))))
f₁(0) -> 0
p₁(s₁(x)) -> x

The set Q consists of the following terms:

f₁(s₁(x0))
f₁(0)
p₁(s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

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

F₁(s₁(x)) -> F₁(p₁(s₁(x)))

The TRS R consists of the following rules:

f₁(s₁(x)) -> s₁(s₁(f₁(p₁(s₁(x)))))
f₁(0) -> 0
p₁(s₁(x)) -> x

The set Q consists of the following terms:

f₁(s₁(x0))
f₁(0)
p₁(s₁(x0))

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