Termination w.r.t. Q proof of /tmp/tmps1fFeO/z015.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(a(a(a(x1))))
b(a(x1)) → a(a(x1))
a(a(x1)) → a(c(b(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(a(a(a(x1))))
b(a(x1)) → a(a(x1))
a(a(x1)) → a(c(b(x1)))

Q is empty.

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

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

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

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(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(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(b(x1)) → b(a(a(a(x1))))
b(a(x1)) → a(a(x1))
a(a(x1)) → a(c(b(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 1 less node.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

B(a(x1)) → A(a(x1))
A(a(x1)) → B(x1)

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

POL( A(x₁) ) = x₁

POL( c(x₁) ) = max{0, -1}

POL( b(x₁) ) = x₁ + 1

POL( B(x₁) ) = x₁

POL( a(x₁) ) = x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

  ↳ QTRS Reverse

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

B(a(x1)) → A(a(x1))
A(a(x1)) → B(x1)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

Q is empty.

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

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

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

a(b(x)) → b(a(a(a(x))))
b(a(x)) → a(a(x))
a(a(x)) → a(c(b(x)))
B(a(x)) → A(a(x))
A(a(x)) → B(x)

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

a(b(x)) → b(a(a(a(x))))
b(a(x)) → a(a(x))
a(a(x)) → a(c(b(x)))
B(a(x)) → A(a(x))
A(a(x)) → B(x)

Q is empty.

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

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

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

a(b(x)) → b(a(a(a(x))))
b(a(x)) → a(a(x))
a(a(x)) → a(c(b(x)))
B(a(x)) → A(a(x))
A(a(x)) → B(x)

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                        ↳ QTRS

                      ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

a(b(x)) → b(a(a(a(x))))
b(a(x)) → a(a(x))
a(a(x)) → a(c(b(x)))
B(a(x)) → A(a(x))
A(a(x)) → B(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:

A¹(B(x)) → A¹(A(x))
B¹(a(x)) → B¹(x)
B¹(a(x)) → A¹(a(b(x)))
B¹(a(x)) → A¹(b(x))
A¹(a(x)) → B¹(c(a(x)))
B¹(a(x)) → A¹(a(a(b(x))))
A¹(b(x)) → A¹(x)
A¹(b(x)) → A¹(a(x))

The TRS R consists of the following rules:

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

A¹(B(x)) → A¹(A(x))
B¹(a(x)) → B¹(x)
B¹(a(x)) → A¹(a(b(x)))
B¹(a(x)) → A¹(b(x))
A¹(a(x)) → B¹(c(a(x)))
B¹(a(x)) → A¹(a(a(b(x))))
A¹(b(x)) → A¹(x)
A¹(b(x)) → A¹(a(x))

The TRS R consists of the following rules:

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → 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 with 5 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                                ↳ UsableRulesProof

                              ↳ QDP

  ↳ QTRS Reverse

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

A¹(b(x)) → A¹(a(x))
A¹(b(x)) → A¹(x)

The TRS R consists of the following rules:

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ RuleRemovalProof

                                ↳ UsableRulesProof

                              ↳ QDP

  ↳ QTRS Reverse

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

A¹(b(x)) → A¹(x)
A¹(b(x)) → A¹(a(x))

The TRS R consists of the following rules:

a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

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:

a(A(x)) → B(x)

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = 1 + 2·x₁
POL(A¹(x₁)) = 2·x₁
POL(B(x₁)) = 1 + 2·x₁
POL(a(x₁)) = 2·x₁
POL(b(x₁)) = 2·x₁
POL(c(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ RuleRemovalProof

                                      ↳ QDP

                                        ↳ UsableRulesReductionPairsProof

                                ↳ UsableRulesProof

                              ↳ QDP

  ↳ QTRS Reverse

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

A¹(b(x)) → A¹(a(x))
A¹(b(x)) → A¹(x)

The TRS R consists of the following rules:

a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(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.

No dependency pairs are removed.

The following rules are removed from R:

a(B(x)) → a(A(x))

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = 1 + x₁
POL(A¹(x₁)) = x₁
POL(B(x₁)) = 2 + x₁
POL(a(x₁)) = 2·x₁
POL(b(x₁)) = 2·x₁
POL(c(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ RuleRemovalProof

                                      ↳ QDP

                                        ↳ UsableRulesReductionPairsProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                ↳ UsableRulesProof

                              ↳ QDP

  ↳ QTRS Reverse

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

A¹(b(x)) → A¹(x)
A¹(b(x)) → A¹(a(x))

The TRS R consists of the following rules:

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

A¹(b(x)) → A¹(x)

Used ordering: POLO with Polynomial interpretation [25]:

POL(A¹(x₁)) = 2·x₁
POL(a(x₁)) = 1 + 2·x₁
POL(b(x₁)) = 1 + 2·x₁
POL(c(x₁)) = x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ RuleRemovalProof

                                      ↳ QDP

                                        ↳ UsableRulesReductionPairsProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                ↳ UsableRulesProof

                              ↳ QDP

  ↳ QTRS Reverse

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

A¹(b(x)) → A¹(a(x))

The TRS R consists of the following rules:

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

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.

A¹(b(x)) → A¹(a(x))

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

POL(A¹(x₁)) = 8·x₁
POL(a(x₁)) = 7 + x₁
POL(b(x₁)) = 8 + x₁
POL(c(x₁)) = 3

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ RuleRemovalProof

                                      ↳ QDP

                                        ↳ UsableRulesReductionPairsProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ PisEmptyProof

                                ↳ UsableRulesProof

                              ↳ QDP

  ↳ QTRS Reverse

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

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

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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                                ↳ UsableRulesProof

                                  ↳ QDP

                              ↳ QDP

  ↳ QTRS Reverse

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

A¹(b(x)) → A¹(x)
A¹(b(x)) → A¹(a(x))

The TRS R consists of the following rules:

a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                                ↳ UsableRulesProof

                                ↳ UsableRulesProof

  ↳ QTRS Reverse

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

B¹(a(x)) → B¹(x)

The TRS R consists of the following rules:

b(a(x)) → a(a(a(b(x))))
a(b(x)) → a(a(x))
a(a(x)) → b(c(a(x)))
a(B(x)) → a(A(x))
a(A(x)) → B(x)

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ UsableRulesReductionPairsProof

                                ↳ UsableRulesProof

  ↳ QTRS Reverse

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

B¹(a(x)) → B¹(x)

R is empty.
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:

B¹(a(x)) → B¹(x)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(B¹(x₁)) = 2·x₁
POL(a(x₁)) = 2·x₁

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ UsableRulesReductionPairsProof

                                      ↳ QDP

                                        ↳ PisEmptyProof

                                ↳ UsableRulesProof

  ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                                ↳ UsableRulesProof

                                ↳ UsableRulesProof

                                  ↳ QDP

  ↳ QTRS Reverse

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

B¹(a(x)) → B¹(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

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

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

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

Q is empty.