Termination w.r.t. Q proof of /tmp/tmpM55sAG/z019.srs

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:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(x1))

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(x1))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(x1))

The set Q is empty.
We have obtained the following QTRS:

b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(a(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(a(x))

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:

C(a(x1)) → C(x1)
A(b(x1)) → B(c(a(x1)))
A(b(x1)) → C(a(x1))
C(a(x1)) → A(c(x1))
B(c(x1)) → B(x1)
B(c(x1)) → B(b(x1))
B(c(x1)) → C(b(b(x1)))
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(x1))

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

  ↳ QTRS Reverse

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

C(a(x1)) → C(x1)
A(b(x1)) → B(c(a(x1)))
A(b(x1)) → C(a(x1))
C(a(x1)) → A(c(x1))
B(c(x1)) → B(x1)
B(c(x1)) → B(b(x1))
B(c(x1)) → C(b(b(x1)))
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(x1))

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:

C(a(x1)) → C(x1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = 2 + 2·x₁
POL(B(x₁)) = 2·x₁
POL(C(x₁)) = 2·x₁
POL(a(x₁)) = 1 + x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

  ↳ QTRS Reverse

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

A(b(x1)) → B(c(a(x1)))
A(b(x1)) → C(a(x1))
C(a(x1)) → A(c(x1))
A(b(x1)) → A(x1)
B(c(x1)) → C(b(b(x1)))
B(c(x1)) → B(b(x1))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(x1))

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.

A(b(x1)) → B(c(a(x1)))
A(b(x1)) → C(a(x1))
A(b(x1)) → A(x1)

The remaining pairs can at least be oriented weakly.

C(a(x1)) → A(c(x1))
B(c(x1)) → C(b(b(x1)))
B(c(x1)) → B(b(x1))
B(c(x1)) → B(x1)

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x₁) ) = x₁

POL( C(x₁) ) = max{0, -1}

POL( c(x₁) ) = max{0, -1}

POL( b(x₁) ) = x₁ + 1

POL( B(x₁) ) = max{0, -1}

POL( a(x₁) ) = x₁

The following usable rules [17] were oriented:

b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(x1))
a(b(x1)) → b(c(a(x1)))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

C(a(x1)) → A(c(x1))
B(c(x1)) → B(x1)
B(c(x1)) → B(b(x1))
B(c(x1)) → C(b(b(x1)))

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(x1))

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 2 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

  ↳ QTRS Reverse

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

B(c(x1)) → B(b(x1))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(c(x1)) → B(b(x1)) at position [0] we obtained the following new rules:

B(c(c(x0))) → B(c(b(b(x0))))

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ SemLabProof

  ↳ QTRS Reverse

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

B(c(c(x0))) → B(c(b(b(x0))))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(x1))

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

B.1(c.1(x1)) → B.1(x1)
B.0(c.0(x1)) → B.0(x1)
B.1(c.1(x1)) → B.0(x1)
B.0(c.0(c.0(x0))) → B.0(c.0(b.0(b.0(x0))))
B.1(c.1(c.1(x0))) → B.0(c.0(b.0(b.1(x0))))

The TRS R consists of the following rules:

a.0(b.0(x1)) → b.1(c.1(a.0(x1)))
b.0(c.0(x1)) → c.0(b.0(b.0(x1)))
c.1(a.0(x1)) → a.0(c.0(x1))
c.1(x0) → c.0(x0)
b.1(x0) → b.0(x0)
b.1(c.1(x1)) → c.0(b.0(b.1(x1)))
a.0(b.1(x1)) → b.1(c.1(a.1(x1)))
c.1(a.1(x1)) → a.1(c.1(x1))
a.1(x0) → a.0(x0)

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

B.1(c.1(x1)) → B.1(x1)
B.0(c.0(x1)) → B.0(x1)
B.1(c.1(x1)) → B.0(x1)
B.0(c.0(c.0(x0))) → B.0(c.0(b.0(b.0(x0))))
B.1(c.1(c.1(x0))) → B.0(c.0(b.0(b.1(x0))))

The TRS R consists of the following rules:

a.0(b.0(x1)) → b.1(c.1(a.0(x1)))
b.0(c.0(x1)) → c.0(b.0(b.0(x1)))
c.1(a.0(x1)) → a.0(c.0(x1))
c.1(x0) → c.0(x0)
b.1(x0) → b.0(x0)
b.1(c.1(x1)) → c.0(b.0(b.1(x1)))
a.0(b.1(x1)) → b.1(c.1(a.1(x1)))
c.1(a.1(x1)) → a.1(c.1(x1))
a.1(x0) → a.0(x0)

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 with 2 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesReductionPairsProof

                              ↳ QDP

  ↳ QTRS Reverse

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

B.0(c.0(x1)) → B.0(x1)
B.0(c.0(c.0(x0))) → B.0(c.0(b.0(b.0(x0))))

The TRS R consists of the following rules:

a.0(b.0(x1)) → b.1(c.1(a.0(x1)))
b.0(c.0(x1)) → c.0(b.0(b.0(x1)))
c.1(a.0(x1)) → a.0(c.0(x1))
c.1(x0) → c.0(x0)
b.1(x0) → b.0(x0)
b.1(c.1(x1)) → c.0(b.0(b.1(x1)))
a.0(b.1(x1)) → b.1(c.1(a.1(x1)))
c.1(a.1(x1)) → a.1(c.1(x1))
a.1(x0) → a.0(x0)

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(B.0(x₁)) = x₁
POL(b.0(x₁)) = x₁
POL(c.0(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesReductionPairsProof

                                  ↳ QDP

                                    ↳ RuleRemovalProof

                              ↳ QDP

  ↳ QTRS Reverse

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

B.0(c.0(x1)) → B.0(x1)
B.0(c.0(c.0(x0))) → B.0(c.0(b.0(b.0(x0))))

The TRS R consists of the following rules:

b.0(c.0(x1)) → c.0(b.0(b.0(x1)))

B.0(c.0(x1)) → B.0(x1)
B.0(c.0(c.0(x0))) → B.0(c.0(b.0(b.0(x0))))

Used ordering: POLO with Polynomial interpretation [25]:

POL(B.0(x₁)) = x₁
POL(b.0(x₁)) = x₁
POL(c.0(x₁)) = 1 + x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesReductionPairsProof

                                  ↳ QDP

                                    ↳ RuleRemovalProof

                                      ↳ QDP

                                        ↳ PisEmptyProof

                              ↳ QDP

  ↳ QTRS Reverse

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

b.0(c.0(x1)) → c.0(b.0(b.0(x1)))

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

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                                ↳ UsableRulesReductionPairsProof

  ↳ QTRS Reverse

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

B.1(c.1(x1)) → B.1(x1)

The TRS R consists of the following rules:

a.0(b.0(x1)) → b.1(c.1(a.0(x1)))
b.0(c.0(x1)) → c.0(b.0(b.0(x1)))
c.1(a.0(x1)) → a.0(c.0(x1))
c.1(x0) → c.0(x0)
b.1(x0) → b.0(x0)
b.1(c.1(x1)) → c.0(b.0(b.1(x1)))
a.0(b.1(x1)) → b.1(c.1(a.1(x1)))
c.1(a.1(x1)) → a.1(c.1(x1))
a.1(x0) → a.0(x0)

B.1(c.1(x1)) → B.1(x1)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(B.1(x₁)) = x₁
POL(c.1(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ SemLabProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                                ↳ UsableRulesReductionPairsProof

                                  ↳ QDP

                                    ↳ PisEmptyProof

  ↳ QTRS Reverse

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.
We have reversed the following QTRS:
The set of rules R is

a(b(x1)) → b(c(a(x1)))
b(c(x1)) → c(b(b(x1)))
c(a(x1)) → a(c(x1))

The set Q is empty.
We have obtained the following QTRS:

b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(a(x))

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

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

b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
a(c(x)) → c(a(x))

Q is empty.