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

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
QDP
      ↳ RuleRemovalProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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


Used ordering: POLO with Polynomial interpretation [25]:

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
QDP
          ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

0(0(0(0(x1)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(1(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 1 SCC with 1 less node.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
QDP
              ↳ Narrowing
  ↳ QTRS Reverse

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

01(0(0(0(x1)))) → 01(1(1(1(x1))))

The TRS R consists of the following rules:

0(0(0(0(x1)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(1(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)))) → 01(1(1(1(x1)))) at position [0] we obtained the following new rules:

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
QDP
                  ↳ QDPToSRSProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

0(0(0(0(x1)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(1(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
QTRS
                      ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
QTRS
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
QTRS
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
QDP
                              ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(1(0(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 4 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
QDP
                                  ↳ RuleRemovalProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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


Used ordering: POLO with Polynomial interpretation [25]:

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
QDP
                                      ↳ RuleRemovalProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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


Used ordering: POLO with Polynomial interpretation [25]:

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
QDP
                                          ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(1(0(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 1 less node.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
QDP
                                              ↳ Narrowing
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(1(0(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)))) → 11(1(1(0(x)))) at position [0] we obtained the following new rules:

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
QDP
                                                  ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(1(0(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 1 less node.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
QDP
                                                      ↳ Narrowing
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule 11(0(0(1(x)))) → 02(1(0(0(x)))) at position [0] we obtained the following new rules:

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ RuleRemovalProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
QDP
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))
1(0(0(0(0(0(01(x))))))) → 0(1(0(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)))) → 0(1(1(1(x1))))
1(0(0(1(x1)))) → 0(0(1(0(x1))))

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.