Termination w.r.t. Q proof of /tmp/tmpx12zcx/3.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(x1)) → a(b(a(x1)))
b(a(b(x1))) → a(c(a(x1)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

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

POL( A(x₁) ) = x₁

POL( c(x₁) ) = 0

POL( b(x₁) ) = x₁

POL( B(x₁) ) = max{0, x₁ - 1}

POL( a(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A(a(x1)) → A(b(a(x1)))

The TRS R consists of the following rules:

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

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

A(a(b(x0))) → A(a(c(a(x0))))
A(a(a(x0))) → A(b(a(b(a(x0)))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A(a(a(x0))) → A(b(a(b(a(x0)))))
A(a(b(x0))) → A(a(c(a(x0))))

The TRS R consists of the following rules:

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A(a(a(x0))) → A(b(a(b(a(x0)))))

The TRS R consists of the following rules:

a(a(x1)) → a(b(a(x1)))
b(a(b(x1))) → a(c(a(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(x0))) → A(b(a(b(a(x0)))))

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( c(x₁) ) = max{0, -1}

POL( b(x₁) ) = 1

POL( a(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ PisEmptyProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

a(a(x1)) → a(b(a(x1)))
b(a(b(x1))) → a(c(a(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.
We have reversed the following QTRS:
The set of rules R is

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

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

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

Q is empty.