Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/RubioSLASHtest4.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(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

f₂(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

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(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

The set Q consists of the following terms:

f₂(a, a)
f₂(a, b)
f₂(s₁(x0), c)
f₂(c, c)

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

F₂(c, c) -> F₂(a, a)
F₂(s₁(X), c) -> F₂(X, c)
F₂(a, b) -> F₂(s₁(a), c)
F₂(a, a) -> F₂(a, b)

The TRS R consists of the following rules:

f₂(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

The set Q consists of the following terms:

f₂(a, a)
f₂(a, b)
f₂(s₁(x0), c)
f₂(c, c)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPAfsSolverProof

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

F₂(c, c) -> F₂(a, a)
F₂(s₁(X), c) -> F₂(X, c)
F₂(a, b) -> F₂(s₁(a), c)
F₂(a, a) -> F₂(a, b)

The TRS R consists of the following rules:

f₂(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

The set Q consists of the following terms:

f₂(a, a)
f₂(a, b)
f₂(s₁(x0), c)
f₂(c, c)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F₂(c, c) -> F₂(a, a)

Used argument filtering: F₂(x1, x2) = x1
c = c
a = a
s₁(x1) = x1
Used ordering: Precedence:

c > a

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

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

F₂(s₁(X), c) -> F₂(X, c)
F₂(a, b) -> F₂(s₁(a), c)
F₂(a, a) -> F₂(a, b)

The TRS R consists of the following rules:

f₂(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

The set Q consists of the following terms:

f₂(a, a)
f₂(a, b)
f₂(s₁(x0), c)
f₂(c, c)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPAfsSolverProof

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

F₂(s₁(X), c) -> F₂(X, c)

The TRS R consists of the following rules:

f₂(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

The set Q consists of the following terms:

f₂(a, a)
f₂(a, b)
f₂(s₁(x0), c)
f₂(c, c)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F₂(s₁(X), c) -> F₂(X, c)

Used argument filtering: F₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPAfsSolverProof

                    ↳ QDP

                      ↳ PisEmptyProof

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

f₂(a, a) -> f₂(a, b)
f₂(a, b) -> f₂(s₁(a), c)
f₂(s₁(X), c) -> f₂(X, c)
f₂(c, c) -> f₂(a, a)

The set Q consists of the following terms:

f₂(a, a)
f₂(a, b)
f₂(s₁(x0), c)
f₂(c, c)

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