Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/RubioSLASHnestrec.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:

f₁(g₁(X)) -> g₁(f₁(f₁(X)))
f₁(h₁(X)) -> h₁(g₁(X))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

f₁(g₁(X)) -> g₁(f₁(f₁(X)))
f₁(h₁(X)) -> h₁(g₁(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(g₁(X)) -> g₁(f₁(f₁(X)))
f₁(h₁(X)) -> h₁(g₁(X))

The set Q consists of the following terms:

f₁(g₁(x0))
f₁(h₁(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(g₁(X)) -> F₁(f₁(X))
F₁(g₁(X)) -> F₁(X)

The TRS R consists of the following rules:

f₁(g₁(X)) -> g₁(f₁(f₁(X)))
f₁(h₁(X)) -> h₁(g₁(X))

The set Q consists of the following terms:

f₁(g₁(x0))
f₁(h₁(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(g₁(X)) -> F₁(f₁(X))
F₁(g₁(X)) -> F₁(X)

The TRS R consists of the following rules:

f₁(g₁(X)) -> g₁(f₁(f₁(X)))
f₁(h₁(X)) -> h₁(g₁(X))

The set Q consists of the following terms:

f₁(g₁(x0))
f₁(h₁(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(g₁(X)) -> F₁(f₁(X))
F₁(g₁(X)) -> F₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F₁(x1) = F₁(x1)
g₁(x1) = g₁(x1)
f₁(x1) = x1
h₁(x1) = h

Lexicographic Path Order [19].
Precedence:

[F₁, g₁, h]

The following usable rules [14] were oriented:

f₁(g₁(X)) -> g₁(f₁(f₁(X)))
f₁(h₁(X)) -> h₁(g₁(X))

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f₁(g₁(X)) -> g₁(f₁(f₁(X)))
f₁(h₁(X)) -> h₁(g₁(X))

The set Q consists of the following terms:

f₁(g₁(x0))
f₁(h₁(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.