Termination w.r.t. Q proof of /tmp/tmptFpCsT/dup06.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(a(b(b(x1)))) → b(b(c(c(a(a(x1))))))
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
b(b(a(a(x1)))) → a(a(c(c(b(b(x1))))))

Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

a(a(b(b(x1)))) → b(b(c(c(a(a(x1))))))
b(b(c(c(x1)))) → c(c(b(b(b(b(x1))))))
b(b(a(a(x1)))) → a(a(c(c(b(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 2 SCCs with 4 less nodes.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

            ↳ UsableRulesProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ QDP

                    ↳ RuleRemovalProof

            ↳ UsableRulesProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

b(b(c(c(x0))))
b(b(a(a(x0))))

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:

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ QDPOrderProof

            ↳ UsableRulesProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

b(b(c(c(x0))))
b(b(a(a(x0))))

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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( c(x₁) ) = x₁ + 1

POL( b(x₁) ) = x₁

POL( B(x₁) ) = max{0, x₁ - 1}

POL( a(x₁) ) = 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

            ↳ UsableRulesProof

          ↳ QDP

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

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

The set Q consists of the following terms:

b(b(c(c(x0))))
b(b(a(a(x0))))

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

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ QDP

                    ↳ MNOCProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

b(b(c(c(x0))))
b(b(a(a(x0))))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ QDP

                    ↳ MNOCProof

                      ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

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

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

The TRS R consists of the following rules:

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

A(a(b(b(x1)))) → A(x1)

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ QDPOrderProof

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

A(a(b(b(x1)))) → A(a(x1))

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x₁) ) = x₁

POL( c(x₁) ) = max{0, -1}

POL( b(x₁) ) = x₁ + 1

POL( a(x₁) ) = x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

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