Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/RubioSLASHkoen.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), X) -> f₂(X, a₁(X))
f₂(X, c₁(X)) -> f₂(s₁(X), X)
f₂(X, X) -> c₁(X)

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

f₂(s₁(X), X) -> f₂(X, a₁(X))
f₂(X, c₁(X)) -> f₂(s₁(X), X)
f₂(X, X) -> c₁(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), X) -> f₂(X, a₁(X))
f₂(X, c₁(X)) -> f₂(s₁(X), X)
f₂(X, X) -> c₁(X)

The set Q consists of the following terms:

f₂(s₁(x0), x0)
f₂(x0, c₁(x0))
f₂(x0, 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:

F₂(X, c₁(X)) -> F₂(s₁(X), X)
F₂(s₁(X), X) -> F₂(X, a₁(X))

The TRS R consists of the following rules:

f₂(s₁(X), X) -> f₂(X, a₁(X))
f₂(X, c₁(X)) -> f₂(s₁(X), X)
f₂(X, X) -> c₁(X)

The set Q consists of the following terms:

f₂(s₁(x0), x0)
f₂(x0, c₁(x0))
f₂(x0, 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:

F₂(X, c₁(X)) -> F₂(s₁(X), X)
F₂(s₁(X), X) -> F₂(X, a₁(X))

The TRS R consists of the following rules:

f₂(s₁(X), X) -> f₂(X, a₁(X))
f₂(X, c₁(X)) -> f₂(s₁(X), X)
f₂(X, X) -> c₁(X)

The set Q consists of the following terms:

f₂(s₁(x0), x0)
f₂(x0, c₁(x0))
f₂(x0, x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.