Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/AG01SLASHHASH3.24.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(0) -> s₁(0)
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> f₁(f₁(s₁(x)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

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

The set Q consists of the following terms:

f₁(0)
f₁(s₁(0))
f₁(s₁(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(s₁(x))) -> F₁(f₁(s₁(x)))
F₁(s₁(s₁(x))) -> F₁(s₁(x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f₁(0)
f₁(s₁(0))
f₁(s₁(s₁(x0)))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

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

F₁(s₁(s₁(x))) -> F₁(f₁(s₁(x)))
F₁(s₁(s₁(x))) -> F₁(s₁(x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f₁(0)
f₁(s₁(0))
f₁(s₁(s₁(x0)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

F₁(s₁(s₁(x))) -> F₁(s₁(x))

The remaining pairs can at least by weakly be oriented.

F₁(s₁(s₁(x))) -> F₁(f₁(s₁(x)))

Used ordering: Combined order from the following AFS and order.
F₁(x1) = x1
s₁(x1) = s₁(x1)
f₁(x1) = f₁(x1)
0 = 0

Lexicographic Path Order [19].
Precedence:

0 > [s₁, f_1]

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

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

F₁(s₁(s₁(x))) -> F₁(f₁(s₁(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f₁(0)
f₁(s₁(0))
f₁(s₁(s₁(x0)))

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