Termination w.r.t. Q proof of /tmp/tmpqEIFas/jambox3.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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(0, a(1, a(x, y))) → B(1, a(0, a(x, y)))
B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, x)
A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))
A(0, a(x, y)) → A(1, a(1, a(x, y)))
A(0, b(0, x)) → A(0, x)
A(0, a(1, a(x, y))) → A(1, a(0, a(x, y)))
A(0, a(x, y)) → A(1, a(x, y))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

B(0, a(1, a(x, y))) → B(1, a(0, a(x, y)))
B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, x)
A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))
A(0, a(x, y)) → A(1, a(1, a(x, y)))
A(0, b(0, x)) → A(0, x)
A(0, a(1, a(x, y))) → A(1, a(0, a(x, y)))
A(0, a(x, y)) → A(1, a(x, y))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

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

B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, x)
A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))
A(0, b(0, x)) → A(0, x)

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, x)
A(0, b(0, x)) → A(0, x)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1
POL(1) = 0
POL(A(x₁, x₂)) = 2 + x₁ + x₂
POL(B(x₁, x₂)) = 1 + x₁ + x₂
POL(a(x₁, x₂)) = 1 + x₁ + x₂
POL(b(x₁, x₂)) = x₁ + x₂

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

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

B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(0, x) → B(0, b(0, x)) at position [1] we obtained the following new rules:

A(0, a(1, a(x0, x1))) → B(0, b(1, a(0, a(x0, x1))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

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

B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, b(0, x)) → B(0, a(0, x))
A(0, a(1, a(x0, x1))) → B(0, b(1, a(0, a(x0, x1))))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

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

B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, b(0, x)) → B(0, a(0, x))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(0, b(0, x)) → B(0, a(0, x)) at position [1] we obtained the following new rules:

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
A(0, b(0, x0)) → B(0, b(0, b(0, x0)))
A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))
A(0, b(0, a(x0, x1))) → B(0, a(1, a(1, a(x0, x1))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))
A(0, b(0, x0)) → B(0, b(0, b(0, x0)))
B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, b(0, a(x0, x1))) → B(0, a(1, a(1, a(x0, x1))))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(0, a(1, a(x, y))) → A(0, a(x, y)) at position [1] we obtained the following new rules:

B(0, a(1, a(0, a(1, a(x0, x1))))) → A(0, a(1, a(0, a(x0, x1))))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))
B(0, a(1, a(0, a(x0, x1)))) → A(0, a(1, a(1, a(x0, x1))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ DependencyGraphProof

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
A(0, b(0, x0)) → B(0, b(0, b(0, x0)))
A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))
A(0, b(0, a(x0, x1))) → B(0, a(1, a(1, a(x0, x1))))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, a(1, a(x0, x1))))) → A(0, a(1, a(0, a(x0, x1))))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))
B(0, a(1, a(0, a(x0, x1)))) → A(0, a(1, a(1, a(x0, x1))))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))
A(0, b(0, x0)) → B(0, b(0, b(0, x0)))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(0, b(0, x0)) → B(0, b(0, b(0, x0))) at position [1] we obtained the following new rules:

A(0, b(0, a(1, a(x0, x1)))) → B(0, b(0, b(1, a(0, a(x0, x1)))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))
A(0, b(0, a(1, a(x0, x1)))) → B(0, b(0, b(1, a(0, a(x0, x1)))))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ QDPApplicativeOrderProof

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].Here, we combined the reduction pair processor with the A-transformation [17] which results in the following intermediate Q-DP Problem.
The a-transformed P is

03(01(1(notProper))) → 02(1(0(notProper)))
03(01(01(x0))) → 02(01(0(x0)))
02(1(0(x0))) → 03(01(01(x0)))
02(1(0(01(x0)))) → 03(01(0(x0)))

The a-transformed usable rules are

01(1(notProper)) → 11(0(notProper))
0(01(x)) → 01(0(x))

The following pairs can be oriented strictly and are deleted.

A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))

The remaining pairs can at least be oriented weakly.

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))

Used ordering: Polynomial interpretation [25]:

POL(0(x₁)) = 0
POL(01(x₁)) = 0
POL(02(x₁)) = x₁
POL(03(x₁)) = 1
POL(1(x₁)) = 1
POL(11(x₁)) = 0
POL(notProper) = 0

The following usable rules [17] were oriented:

b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, b(0, x)) → b(0, a(0, x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ QDPApplicativeOrderProof

                                            ↳ QDP

                                              ↳ Narrowing

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0))) at position [1] we obtained the following new rules:

B(0, a(1, a(0, a(1, a(x0, x1))))) → A(0, b(0, b(1, a(0, a(x0, x1)))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ QDPApplicativeOrderProof

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
B(0, a(1, a(0, a(1, a(x0, x1))))) → A(0, b(0, b(1, a(0, a(x0, x1)))))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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

          ↳ RuleRemovalProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ Narrowing

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ QDPApplicativeOrderProof

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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