Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/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)

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(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

          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

F₂(c, c) -> F₂(a, a)

The remaining pairs can at least by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
F₂(x1, x2) = x1
c = c
a = a
s₁(x1) = x1
b = b

Lexicographic Path Order [19].
Precedence:

c > [a, b]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ 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 [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

F₂(s₁(X), c) -> F₂(X, c)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F₂(x1, x2) = F₁(x1)
s₁(x1) = s₁(x1)
c = c

Lexicographic Path Order [19].
Precedence:

[s₁, c] > F₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ 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.