Termination w.r.t. Q proof of /tmp/tmpxUKHqc/07.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

  ↳ QTRS Reverse

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

A(a(a(x1))) → A(b(b(x1)))

The TRS R consists of the following rules:

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

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

POL( A(x₁) ) = max{0, x₁ - 1}

POL( a(x₁) ) = x₁ + 1

POL( b(x₁) ) = max{0, -1}

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

  ↳ QTRS Reverse

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

B(a(x1)) → B(b(x1))

The remaining pairs can at least be oriented weakly.

B(a(b(x1))) → B(a(a(x1)))

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

POL( B(x₁) ) = x₁ + 1

POL( a(x₁) ) = 1

POL( b(x₁) ) = 0

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ Narrowing

  ↳ QTRS Reverse

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

B(a(b(x1))) → B(a(a(x1)))

The TRS R consists of the following rules:

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

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

B(a(b(a(a(x0))))) → B(a(a(b(b(x0)))))
B(a(b(a(x0)))) → B(a(b(b(x0))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ SemLabProof

  ↳ QTRS Reverse

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

B(a(b(a(a(x0))))) → B(a(a(b(b(x0)))))
B(a(b(a(x0)))) → B(a(b(b(x0))))

The TRS R consists of the following rules:

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

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.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(a.0(b.1(a.1(a.1(x0))))) → B.1(a.1(a.0(b.0(b.1(x0)))))
B.1(a.0(b.1(a.1(a.0(x0))))) → B.1(a.1(a.0(b.0(b.0(x0)))))
B.1(a.0(b.1(a.0(x0)))) → B.1(a.0(b.0(b.0(x0))))
B.1(a.0(b.1(a.1(x0)))) → B.1(a.0(b.0(b.1(x0))))

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ SemLabProof

                          ↳ QDP

                            ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

B.1(a.0(b.1(a.1(a.1(x0))))) → B.1(a.1(a.0(b.0(b.1(x0)))))
B.1(a.0(b.1(a.1(a.0(x0))))) → B.1(a.1(a.0(b.0(b.0(x0)))))
B.1(a.0(b.1(a.0(x0)))) → B.1(a.0(b.0(b.0(x0))))
B.1(a.0(b.1(a.1(x0)))) → B.1(a.0(b.0(b.1(x0))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 4 less nodes.
We have reversed the following QTRS:
The set of rules R is

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

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

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

Q is empty.