Termination w.r.t. Q proof of /tmp/tmp5Q6-hX/z21.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(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(f(a, b), x) → F(a, x)
F(x, f(y, z)) → F(x, y)
F(f(b, a), x) → F(b, f(b, x))
F(f(a, b), x) → F(a, f(a, x))
F(f(b, a), x) → F(b, x)
F(x, f(y, z)) → F(f(x, y), z)

The TRS R consists of the following rules:

f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

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

F(f(a, b), x) → F(a, x)
F(x, f(y, z)) → F(x, y)
F(f(b, a), x) → F(b, f(b, x))
F(f(a, b), x) → F(a, f(a, x))
F(f(b, a), x) → F(b, x)
F(x, f(y, z)) → F(f(x, y), z)

The TRS R consists of the following rules:

f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

F(f(a, b), x) → F(a, x)
F(x, f(y, z)) → F(x, y)
F(f(b, a), x) → F(b, x)

Used ordering: POLO with Polynomial interpretation [25]:

POL(F(x₁, x₂)) = 2·x₁ + 2·x₂
POL(a) = 2
POL(b) = 2
POL(f(x₁, x₂)) = 2 + x₁ + x₂

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ SemLabProof

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

F(f(b, a), x) → F(b, f(b, x))
F(f(a, b), x) → F(a, f(a, x))
F(x, f(y, z)) → F(f(x, y), z)

The TRS R consists of the following rules:

f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.a: 0
f: x₀
b: 1
F: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

F.1-0(x, f.0-1(y, z)) → F.1-1(f.1-0(x, y), z)
F.0-1(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.1-0(f.1-0(b., a.), x) → F.1-1(b., f.1-0(b., x))
F.0-0(f.0-1(a., b.), x) → F.0-0(a., f.0-0(a., x))
F.1-1(x, f.1-0(y, z)) → F.1-0(f.1-1(x, y), z)
F.0-1(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
F.1-0(x, f.0-0(y, z)) → F.1-0(f.1-0(x, y), z)
F.1-1(f.1-0(b., a.), x) → F.1-1(b., f.1-1(b., x))
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.0-1(f.0-1(a., b.), x) → F.0-0(a., f.0-1(a., x))
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)

The TRS R consists of the following rules:

f.0-1(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.1-0(f.1-0(b., a.), x) → f.1-1(b., f.1-0(b., x))
f.0-1(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.0-1(f.0-1(a., b.), x) → f.0-0(a., f.0-1(a., x))
f.1-0(x, f.0-1(y, z)) → f.1-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.0-1(a., b.), x) → f.0-0(a., f.0-0(a., x))
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)
f.1-0(x, f.0-0(y, z)) → f.1-0(f.1-0(x, y), z)
f.1-1(f.1-0(b., a.), x) → f.1-1(b., f.1-1(b., x))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

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

F.1-0(x, f.0-1(y, z)) → F.1-1(f.1-0(x, y), z)
F.0-1(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.1-0(f.1-0(b., a.), x) → F.1-1(b., f.1-0(b., x))
F.0-0(f.0-1(a., b.), x) → F.0-0(a., f.0-0(a., x))
F.1-1(x, f.1-0(y, z)) → F.1-0(f.1-1(x, y), z)
F.0-1(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
F.1-0(x, f.0-0(y, z)) → F.1-0(f.1-0(x, y), z)
F.1-1(f.1-0(b., a.), x) → F.1-1(b., f.1-1(b., x))
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.0-1(f.0-1(a., b.), x) → F.0-0(a., f.0-1(a., x))
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)

The TRS R consists of the following rules:

f.0-1(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.1-0(f.1-0(b., a.), x) → f.1-1(b., f.1-0(b., x))
f.0-1(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.0-1(f.0-1(a., b.), x) → f.0-0(a., f.0-1(a., x))
f.1-0(x, f.0-1(y, z)) → f.1-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.0-1(a., b.), x) → f.0-0(a., f.0-0(a., x))
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)
f.1-0(x, f.0-0(y, z)) → f.1-0(f.1-0(x, y), z)
f.1-1(f.1-0(b., a.), x) → f.1-1(b., f.1-1(b., x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                  ↳ QDP

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

F.0-1(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.0-0(f.0-1(a., b.), x) → F.0-0(a., f.0-0(a., x))
F.0-1(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.0-1(f.0-1(a., b.), x) → F.0-0(a., f.0-1(a., x))
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)

The TRS R consists of the following rules:

f.0-1(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.1-0(f.1-0(b., a.), x) → f.1-1(b., f.1-0(b., x))
f.0-1(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.0-1(f.0-1(a., b.), x) → f.0-0(a., f.0-1(a., x))
f.1-0(x, f.0-1(y, z)) → f.1-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.0-1(a., b.), x) → f.0-0(a., f.0-0(a., x))
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)
f.1-0(x, f.0-0(y, z)) → f.1-0(f.1-0(x, y), z)
f.1-1(f.1-0(b., a.), x) → f.1-1(b., f.1-1(b., x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

F.0-1(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.0-0(f.0-1(a., b.), x) → F.0-0(a., f.0-0(a., x))
F.0-1(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
F.0-1(f.0-1(a., b.), x) → F.0-0(a., f.0-1(a., x))

The following rules are removed from R:

f.0-1(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-1(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.0-1(f.0-1(a., b.), x) → f.0-0(a., f.0-1(a., x))
f.0-0(f.0-1(a., b.), x) → f.0-0(a., f.0-0(a., x))

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0-0(x₁, x₂)) = 1 + x₁ + x₂
POL(F.0-1(x₁, x₂)) = 1 + x₁ + x₂
POL(a.) = 0
POL(b.) = 1
POL(f.0-0(x₁, x₂)) = x₁ + x₂
POL(f.0-1(x₁, x₂)) = x₁ + x₂
POL(f.1-0(x₁, x₂)) = 1 + x₁ + x₂
POL(f.1-1(x₁, x₂)) = x₁ + x₂

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                  ↳ QDP

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

F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)

The TRS R consists of the following rules:

f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                  ↳ QDP

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

F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)

The TRS R consists of the following rules:

f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(F.0-0(x₁, x₂)) = x₂
POL(f.0-0(x₁, x₂)) = 1 + x₁ + x₂
POL(f.0-1(x₁, x₂)) = 0

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ PisEmptyProof

                  ↳ QDP

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

f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

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

F.1-0(x, f.0-1(y, z)) → F.1-1(f.1-0(x, y), z)
F.1-0(f.1-0(b., a.), x) → F.1-1(b., f.1-0(b., x))
F.1-1(x, f.1-0(y, z)) → F.1-0(f.1-1(x, y), z)
F.1-0(x, f.0-0(y, z)) → F.1-0(f.1-0(x, y), z)
F.1-1(f.1-0(b., a.), x) → F.1-1(b., f.1-1(b., x))
F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)

The TRS R consists of the following rules:

f.0-1(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.1-0(f.1-0(b., a.), x) → f.1-1(b., f.1-0(b., x))
f.0-1(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.0-1(f.0-1(a., b.), x) → f.0-0(a., f.0-1(a., x))
f.1-0(x, f.0-1(y, z)) → f.1-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.0-1(a., b.), x) → f.0-0(a., f.0-0(a., x))
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)
f.1-0(x, f.0-0(y, z)) → f.1-0(f.1-0(x, y), z)
f.1-1(f.1-0(b., a.), x) → f.1-1(b., f.1-1(b., x))

F.1-0(x, f.0-1(y, z)) → F.1-1(f.1-0(x, y), z)
F.1-0(f.1-0(b., a.), x) → F.1-1(b., f.1-0(b., x))
F.1-0(x, f.0-0(y, z)) → F.1-0(f.1-0(x, y), z)
F.1-1(f.1-0(b., a.), x) → F.1-1(b., f.1-1(b., x))

The following rules are removed from R:

f.1-0(f.1-0(b., a.), x) → f.1-1(b., f.1-0(b., x))
f.1-0(x, f.0-0(y, z)) → f.1-0(f.1-0(x, y), z)
f.1-1(f.1-0(b., a.), x) → f.1-1(b., f.1-1(b., x))
f.1-0(x, f.0-1(y, z)) → f.1-1(f.1-0(x, y), z)

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.1-0(x₁, x₂)) = 1 + x₁ + x₂
POL(F.1-1(x₁, x₂)) = 1 + x₁ + x₂
POL(a.) = 1
POL(b.) = 0
POL(f.0-0(x₁, x₂)) = 1 + x₁ + x₂
POL(f.0-1(x₁, x₂)) = 1 + x₁ + x₂
POL(f.1-0(x₁, x₂)) = x₁ + x₂
POL(f.1-1(x₁, x₂)) = x₁ + x₂

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

F.1-1(x, f.1-0(y, z)) → F.1-0(f.1-1(x, y), z)
F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)

The TRS R consists of the following rules:

f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

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

F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)

The TRS R consists of the following rules:

f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(F.1-1(x₁, x₂)) = x₂
POL(f.1-0(x₁, x₂)) = 0
POL(f.1-1(x₁, x₂)) = 1 + x₁ + x₂

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ PisEmptyProof

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

f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)

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