Termination w.r.t. Q proof of ../tpdb/SRS/Gebhardt/20.srs

MAYBE Termination w.r.t. Q proof of ../tpdb/SRS/Gebhardt/20.srs
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:

0(0(0(0(x1)))) → 1(0(1(1(x1))))
1(0(0(1(x1)))) → 0(1(0(0(x1))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

0(0(0(0(x1)))) → 1(0(1(1(x1))))
1(0(0(1(x1)))) → 0(1(0(0(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:

0¹(0(0(0(x1)))) → 1¹(x1)
0¹(0(0(0(x1)))) → 1¹(1(x1))
1¹(0(0(1(x1)))) → 1¹(0(0(x1)))
1¹(0(0(1(x1)))) → 0¹(0(x1))
0¹(0(0(0(x1)))) → 1¹(0(1(1(x1))))
1¹(0(0(1(x1)))) → 0¹(x1)
1¹(0(0(1(x1)))) → 0¹(1(0(0(x1))))
0¹(0(0(0(x1)))) → 0¹(1(1(x1)))

The TRS R consists of the following rules:

0(0(0(0(x1)))) → 1(0(1(1(x1))))
1(0(0(1(x1)))) → 0(1(0(0(x1))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

0¹(0(0(0(x1)))) → 1¹(x1)
0¹(0(0(0(x1)))) → 1¹(1(x1))
1¹(0(0(1(x1)))) → 1¹(0(0(x1)))
1¹(0(0(1(x1)))) → 0¹(0(x1))
0¹(0(0(0(x1)))) → 1¹(0(1(1(x1))))
1¹(0(0(1(x1)))) → 0¹(x1)
1¹(0(0(1(x1)))) → 0¹(1(0(0(x1))))
0¹(0(0(0(x1)))) → 0¹(1(1(x1)))

The TRS R consists of the following rules:

0(0(0(0(x1)))) → 1(0(1(1(x1))))
1(0(0(1(x1)))) → 0(1(0(0(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.

0¹(0(0(0(x1)))) → 1¹(x1)
0¹(0(0(0(x1)))) → 1¹(1(x1))
1¹(0(0(1(x1)))) → 1¹(0(0(x1)))
1¹(0(0(1(x1)))) → 0¹(0(x1))
1¹(0(0(1(x1)))) → 0¹(x1)
0¹(0(0(0(x1)))) → 0¹(1(1(x1)))

The remaining pairs can at least be oriented weakly.

0¹(0(0(0(x1)))) → 1¹(0(1(1(x1))))
1¹(0(0(1(x1)))) → 0¹(1(0(0(x1))))

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

POL( 0¹(x₁) ) = x₁ + 1

POL( 0(x₁) ) = x₁ + 1

POL( 1(x₁) ) = x₁ + 1

POL( 1¹(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

0(0(0(0(x1)))) → 1(0(1(1(x1))))
1(0(0(1(x1)))) → 0(1(0(0(x1))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

0¹(0(0(0(x1)))) → 1¹(0(1(1(x1))))
1¹(0(0(1(x1)))) → 0¹(1(0(0(x1))))

The TRS R consists of the following rules:

0(0(0(0(x1)))) → 1(0(1(1(x1))))
1(0(0(1(x1)))) → 0(1(0(0(x1))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 0¹(0(0(0(x1)))) → 1¹(0(1(1(x1)))) at position [0] we obtained the following new rules:

0¹(0(0(0(0(0(1(x0))))))) → 1¹(0(1(0(1(0(0(x0)))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

0¹(0(0(0(0(0(1(x0))))))) → 1¹(0(1(0(1(0(0(x0)))))))
1¹(0(0(1(x1)))) → 0¹(1(0(0(x1))))

The TRS R consists of the following rules:

0(0(0(0(x1)))) → 1(0(1(1(x1))))
1(0(0(1(x1)))) → 0(1(0(0(x1))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 1¹(0(0(1(x1)))) → 0¹(1(0(0(x1)))) at position [0] we obtained the following new rules:

1¹(0(0(1(1(x0))))) → 0¹(0(1(0(0(x0)))))
1¹(0(0(1(0(0(x0)))))) → 0¹(1(1(0(1(1(x0))))))
1¹(0(0(1(0(0(0(x0))))))) → 0¹(1(0(1(0(1(1(x0)))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

1¹(0(0(1(0(0(0(x0))))))) → 0¹(1(0(1(0(1(1(x0)))))))
0¹(0(0(0(0(0(1(x0))))))) → 1¹(0(1(0(1(0(0(x0)))))))
1¹(0(0(1(1(x0))))) → 0¹(0(1(0(0(x0)))))
1¹(0(0(1(0(0(x0)))))) → 0¹(1(1(0(1(1(x0))))))

The TRS R consists of the following rules:

0(0(0(0(x1)))) → 1(0(1(1(x1))))
1(0(0(1(x1)))) → 0(1(0(0(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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

0(0(0(0(x1)))) → 1(0(1(1(x1))))
1(0(0(1(x1)))) → 0(1(0(0(x1))))
1¹(0(0(1(0(0(0(x0))))))) → 0¹(1(0(1(0(1(1(x0)))))))
0¹(0(0(0(0(0(1(x0))))))) → 1¹(0(1(0(1(0(0(x0)))))))
1¹(0(0(1(1(x0))))) → 0¹(0(1(0(0(x0)))))
1¹(0(0(1(0(0(x0)))))) → 0¹(1(1(0(1(1(x0))))))

Q is empty.

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

0(0(0(0(x1)))) → 1(0(1(1(x1))))
1(0(0(1(x1)))) → 0(1(0(0(x1))))
1¹(0(0(1(0(0(0(x0))))))) → 0¹(1(0(1(0(1(1(x0)))))))
0¹(0(0(0(0(0(1(x0))))))) → 1¹(0(1(0(1(0(0(x0)))))))
1¹(0(0(1(1(x0))))) → 0¹(0(1(0(0(x0)))))
1¹(0(0(1(0(0(x0)))))) → 0¹(1(1(0(1(1(x0))))))

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                            ↳ QTRS

                          ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

0²(0(0(1(0(0(1¹(x))))))) → 1²(0(1(0(1(0¹(x))))))
0²(0(0(1(0(0(1¹(x))))))) → 0²(1(0(1(0¹(x)))))
0²(0(0(1(0(0(1¹(x))))))) → 1²(1(0(1(0(1(0¹(x)))))))
0²(0(1(0(0(1¹(x)))))) → 1²(1(0¹(x)))
1²(0(0(0(0(0(0¹(x))))))) → 1²(0(1¹(x)))
0²(0(1(0(0(1¹(x)))))) → 1²(1(0(1(1(0¹(x))))))
0²(0(0(1(0(0(1¹(x))))))) → 0²(1(0¹(x)))
1²(0(0(0(0(0(0¹(x))))))) → 0²(0(1(0(1(0(1¹(x)))))))
1²(1(0(0(1¹(x))))) → 0²(0¹(x))
1²(0(0(0(0(0(0¹(x))))))) → 0²(1(0(1(0(1¹(x))))))
0²(0(1(0(0(1¹(x)))))) → 1²(0(1(1(0¹(x)))))
1²(1(0(0(1¹(x))))) → 0²(0(1(0(0¹(x)))))
1²(0(0(1(x)))) → 0²(1(0(x)))
0²(0(0(0(x)))) → 1²(x)
0²(0(0(1(0(0(1¹(x))))))) → 1²(0(1(0¹(x))))
1²(0(0(0(0(0(0¹(x))))))) → 0²(1(0(1¹(x))))
1²(0(0(0(0(0(0¹(x))))))) → 1²(0(1(0(1¹(x)))))
0²(0(0(0(x)))) → 0²(1(x))
1²(0(0(1(x)))) → 0²(x)
0²(0(0(1(0(0(1¹(x))))))) → 1²(0¹(x))
1²(0(0(0(0(0(0¹(x))))))) → 0²(1¹(x))
0²(0(1(0(0(1¹(x)))))) → 0²(1(1(0¹(x))))
0²(0(0(0(x)))) → 1²(0(1(x)))
1²(0(0(1(x)))) → 0²(0(1(0(x))))
0²(0(0(0(x)))) → 1²(1(0(1(x))))
1²(1(0(0(1¹(x))))) → 1²(0(0¹(x)))
1²(0(0(1(x)))) → 1²(0(x))
1²(1(0(0(1¹(x))))) → 0²(1(0(0¹(x))))
0²(0(1(0(0(1¹(x)))))) → 1²(0¹(x))

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

0²(0(0(1(0(0(1¹(x))))))) → 1²(0(1(0(1(0¹(x))))))
0²(0(0(1(0(0(1¹(x))))))) → 0²(1(0(1(0¹(x)))))
0²(0(0(1(0(0(1¹(x))))))) → 1²(1(0(1(0(1(0¹(x)))))))
0²(0(1(0(0(1¹(x)))))) → 1²(1(0¹(x)))
1²(0(0(0(0(0(0¹(x))))))) → 1²(0(1¹(x)))
0²(0(1(0(0(1¹(x)))))) → 1²(1(0(1(1(0¹(x))))))
0²(0(0(1(0(0(1¹(x))))))) → 0²(1(0¹(x)))
1²(0(0(0(0(0(0¹(x))))))) → 0²(0(1(0(1(0(1¹(x)))))))
1²(1(0(0(1¹(x))))) → 0²(0¹(x))
1²(0(0(0(0(0(0¹(x))))))) → 0²(1(0(1(0(1¹(x))))))
0²(0(1(0(0(1¹(x)))))) → 1²(0(1(1(0¹(x)))))
1²(1(0(0(1¹(x))))) → 0²(0(1(0(0¹(x)))))
1²(0(0(1(x)))) → 0²(1(0(x)))
0²(0(0(0(x)))) → 1²(x)
0²(0(0(1(0(0(1¹(x))))))) → 1²(0(1(0¹(x))))
1²(0(0(0(0(0(0¹(x))))))) → 0²(1(0(1¹(x))))
1²(0(0(0(0(0(0¹(x))))))) → 1²(0(1(0(1¹(x)))))
0²(0(0(0(x)))) → 0²(1(x))
1²(0(0(1(x)))) → 0²(x)
0²(0(0(1(0(0(1¹(x))))))) → 1²(0¹(x))
1²(0(0(0(0(0(0¹(x))))))) → 0²(1¹(x))
0²(0(1(0(0(1¹(x)))))) → 0²(1(1(0¹(x))))
0²(0(0(0(x)))) → 1²(0(1(x)))
1²(0(0(1(x)))) → 0²(0(1(0(x))))
0²(0(0(0(x)))) → 1²(1(0(1(x))))
1²(1(0(0(1¹(x))))) → 1²(0(0¹(x)))
1²(0(0(1(x)))) → 1²(0(x))
1²(1(0(0(1¹(x))))) → 0²(1(0(0¹(x))))
0²(0(1(0(0(1¹(x)))))) → 1²(0¹(x))

The TRS R consists of the following rules:

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

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 21 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

0²(0(0(0(x)))) → 1²(1(0(1(x))))
0²(0(0(0(x)))) → 0²(1(x))
1²(0(0(1(x)))) → 0²(x)
1²(0(0(1(x)))) → 1²(0(x))
0²(0(0(0(x)))) → 1²(0(1(x)))
1²(0(0(1(x)))) → 0²(0(1(0(x))))
0²(0(0(0(x)))) → 1²(x)
1²(0(0(1(x)))) → 0²(1(0(x)))

The TRS R consists of the following rules:

0(0(0(0(x)))) → 1(1(0(1(x))))
1(0(0(1(x)))) → 0(0(1(0(x))))
0(0(0(1(0(0(1¹(x))))))) → 1(1(0(1(0(1(0¹(x)))))))
1(0(0(0(0(0(0¹(x))))))) → 0(0(1(0(1(0(1¹(x)))))))
1(1(0(0(1¹(x))))) → 0(0(1(0(0¹(x)))))
0(0(1(0(0(1¹(x)))))) → 1(1(0(1(1(0¹(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 dependency pairs:

0²(0(0(0(x)))) → 0²(1(x))
1²(0(0(1(x)))) → 0²(x)
1²(0(0(1(x)))) → 1²(0(x))
0²(0(0(0(x)))) → 1²(0(1(x)))
0²(0(0(0(x)))) → 1²(x)
1²(0(0(1(x)))) → 0²(1(0(x)))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0(x₁)) = 1 + 2·x₁
POL(0¹(x₁)) = x₁
POL(0²(x₁)) = 2·x₁
POL(1(x₁)) = 1 + 2·x₁
POL(1¹(x₁)) = x₁
POL(1²(x₁)) = 2·x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

                                    ↳ QDP

                                      ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

0²(0(0(0(x)))) → 1²(1(0(1(x))))
1²(0(0(1(x)))) → 0²(0(1(0(x))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 0²(0(0(0(x)))) → 1²(1(0(1(x)))) at position [0] we obtained the following new rules:

0²(0(0(0(0(0(1(x0))))))) → 1²(1(0(0(0(1(0(x0)))))))
0²(0(0(0(0(0(0(0(0(0¹(x0)))))))))) → 1²(1(0(0(0(1(0(1(0(1¹(x0))))))))))
0²(0(0(0(1(0(0(1¹(x0)))))))) → 1²(1(0(0(0(1(0(0¹(x0))))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

0²(0(0(0(1(0(0(1¹(x0)))))))) → 1²(1(0(0(0(1(0(0¹(x0))))))))
0²(0(0(0(0(0(1(x0))))))) → 1²(1(0(0(0(1(0(x0)))))))
0²(0(0(0(0(0(0(0(0(0¹(x0)))))))))) → 1²(1(0(0(0(1(0(1(0(1¹(x0))))))))))
1²(0(0(1(x)))) → 0²(0(1(0(x))))

The TRS R consists of the following rules:

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

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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

0²(0(0(0(0(0(1(x0))))))) → 1²(1(0(0(0(1(0(x0)))))))
1²(0(0(1(x)))) → 0²(0(1(0(x))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 1²(0(0(1(x)))) → 0²(0(1(0(x)))) at position [0] we obtained the following new rules:

1²(0(0(1(0(0(1(0(0(1¹(x0)))))))))) → 0²(0(1(1(1(0(1(0(1(0¹(x0))))))))))
1²(0(0(1(0(1(x0)))))) → 0²(0(0(0(1(0(x0))))))
1²(0(0(1(0(0(0(0(0¹(x0))))))))) → 0²(0(0(0(1(0(1(0(1¹(x0)))))))))
1²(0(0(1(0(1(0(0(1¹(x0))))))))) → 0²(0(1(1(1(0(1(1(0¹(x0)))))))))
1²(0(0(1(0(0(0(x0))))))) → 0²(0(1(1(1(0(1(x0)))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

1²(0(0(1(0(0(1(0(0(1¹(x0)))))))))) → 0²(0(1(1(1(0(1(0(1(0¹(x0))))))))))
1²(0(0(1(0(1(x0)))))) → 0²(0(0(0(1(0(x0))))))
1²(0(0(1(0(0(0(0(0¹(x0))))))))) → 0²(0(0(0(1(0(1(0(1¹(x0)))))))))
1²(0(0(1(0(1(0(0(1¹(x0))))))))) → 0²(0(1(1(1(0(1(1(0¹(x0)))))))))
0²(0(0(0(0(0(1(x0))))))) → 1²(1(0(0(0(1(0(x0)))))))
1²(0(0(1(0(0(0(x0))))))) → 0²(0(1(1(1(0(1(x0)))))))

The TRS R consists of the following rules:

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

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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ RuleRemovalProof

                                    ↳ QDP

                                      ↳ Narrowing

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

1²(0(0(1(0(1(x0)))))) → 0²(0(0(0(1(0(x0))))))
0²(0(0(0(0(0(1(x0))))))) → 1²(1(0(0(0(1(0(x0)))))))
1²(0(0(1(0(0(0(x0))))))) → 0²(0(1(1(1(0(1(x0)))))))

The TRS R consists of the following rules:

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

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

0(0(0(0(x1)))) → 1(0(1(1(x1))))
1(0(0(1(x1)))) → 0(1(0(0(x1))))

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

0(0(0(0(x)))) → 1(1(0(1(x))))
1(0(0(1(x)))) → 0(0(1(0(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:

0(0(0(0(x)))) → 1(1(0(1(x))))
1(0(0(1(x)))) → 0(0(1(0(x))))

Q is empty.

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

0(0(0(0(x1)))) → 1(0(1(1(x1))))
1(0(0(1(x1)))) → 0(1(0(0(x1))))

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

0(0(0(0(x)))) → 1(1(0(1(x))))
1(0(0(1(x)))) → 0(0(1(0(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:

0(0(0(0(x)))) → 1(1(0(1(x))))
1(0(0(1(x)))) → 0(0(1(0(x))))

Q is empty.