Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/RubioSLASHmfp95.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

f₂(s₁(X), Y) -> h₁(s₁(f₂(h₁(Y), X)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

f₂(s₁(X), Y) -> h₁(s₁(f₂(h₁(Y), 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:

f₂(s₁(X), Y) -> h₁(s₁(f₂(h₁(Y), X)))

The set Q consists of the following terms:

f₂(s₁(x0), x1)

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

F₂(s₁(X), Y) -> F₂(h₁(Y), X)

The TRS R consists of the following rules:

f₂(s₁(X), Y) -> h₁(s₁(f₂(h₁(Y), X)))

The set Q consists of the following terms:

f₂(s₁(x0), x1)

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:

F₂(s₁(X), Y) -> F₂(h₁(Y), X)

The TRS R consists of the following rules:

f₂(s₁(X), Y) -> h₁(s₁(f₂(h₁(Y), X)))

The set Q consists of the following terms:

f₂(s₁(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.