Termination w.r.t. Q proof of /tmp/tmprHbzna/teparla2.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(y, f(x, f(a, x))) → f(f(f(a, x), f(x, a)), f(a, y))
f(x, f(x, y)) → f(f(f(x, a), a), a)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ SemLabProof

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

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

The TRS R consists of the following rules:

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

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: 1
f: 0
F: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

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 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesReductionPairsProof

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

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

The TRS R consists of the following rules:

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

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.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesReductionPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesReductionPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ RuleRemovalProof

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

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

The TRS R consists of the following rules:

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

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 rules of the TRS R:

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

Used ordering: POLO with Polynomial interpretation [25]:

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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesReductionPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ UsableRulesReductionPairsProof

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesReductionPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ UsableRulesReductionPairsProof

                                ↳ QDP

                                  ↳ QDPOrderProof

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

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

R is empty.
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(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.0-0(f.1-1(a., x), f.1-1(x, a.)), f.1-0(a., y))

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

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

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesReductionPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ RuleRemovalProof

                            ↳ QDP

                              ↳ UsableRulesReductionPairsProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
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.