Termination w.r.t. Q proof of TRS/Rubiotest4.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:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

The signature Sigma is {f}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

The set Q consists of the following terms:

f(a, a)
f(a, b)
f(s(x0), c)
f(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:

F(c, c) → F(a, a)
F(s(X), c) → F(X, c)
F(a, b) → F(s(a), c)
F(a, a) → F(a, b)

The TRS R consists of the following rules:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

The set Q consists of the following terms:

f(a, a)
f(a, b)
f(s(x0), c)
f(c, c)

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

F(c, c) → F(a, a)
F(s(X), c) → F(X, c)
F(a, b) → F(s(a), c)
F(a, a) → F(a, b)

The TRS R consists of the following rules:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

The set Q consists of the following terms:

f(a, a)
f(a, b)
f(s(x0), c)
f(c, c)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ QDPOrderProof

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

F(c, c) → F(a, a)
F(s(X), c) → F(X, c)
F(a, b) → F(s(a), c)
F(a, a) → F(a, b)

The TRS R consists of the following rules:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

The set Q consists of the following terms:

f(a, a)
f(a, b)
f(s(x0), c)
f(c, c)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

F(c, c) → F(a, a)

The remaining pairs can at least be oriented weakly.

F(s(X), c) → F(X, c)
F(a, b) → F(s(a), c)
F(a, a) → F(a, b)

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

Lexicographic path order with status [19].
Precedence:

F₁ > c > a
F₁ > b > a

Status:

b: multiset
a: multiset
c: multiset
F₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

F(s(X), c) → F(X, c)
F(a, b) → F(s(a), c)
F(a, a) → F(a, b)

The TRS R consists of the following rules:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

The set Q consists of the following terms:

f(a, a)
f(a, b)
f(s(x0), c)
f(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

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

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

F(s(X), c) → F(X, c)

The TRS R consists of the following rules:

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

The set Q consists of the following terms:

f(a, a)
f(a, b)
f(s(x0), c)
f(c, c)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

F(s(X), c) → F(X, c)

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

Lexicographic path order with status [19].
Precedence:

c > F₁

Status:

c: multiset
s₁: multiset
F₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ PisEmptyProof

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

f(a, a) → f(a, b)
f(a, b) → f(s(a), c)
f(s(X), c) → f(X, c)
f(c, c) → f(a, a)

The set Q consists of the following terms:

f(a, a)
f(a, b)
f(s(x0), c)
f(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.