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:

f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)

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:

F(f(a, b), x) → F(a, x)
F(x, f(y, z)) → F(x, y)
F(f(b, a), x) → F(b, f(b, x))
F(f(a, b), x) → F(a, f(a, x))
F(f(b, a), x) → F(b, x)
F(x, f(y, z)) → F(f(x, y), z)

The TRS R consists of the following rules:

f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ RuleRemovalProof

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

F(f(a, b), x) → F(a, x)
F(x, f(y, z)) → F(x, y)
F(f(b, a), x) → F(b, f(b, x))
F(f(a, b), x) → F(a, f(a, x))
F(f(b, a), x) → F(b, x)
F(x, f(y, z)) → F(f(x, y), z)

The TRS R consists of the following rules:

f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)

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:

F(f(a, b), x) → F(a, x)
F(x, f(y, z)) → F(x, y)
F(f(b, a), x) → F(b, x)


Used ordering: POLO with Polynomial interpretation [25]:

POL(F(x1, x2)) = 2·x1 + 2·x2   
POL(a) = 2   
POL(b) = 2   
POL(f(x1, x2)) = 2 + x1 + x2   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
QDP
          ↳ SemLabProof

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

F(f(b, a), x) → F(b, f(b, x))
F(f(a, b), x) → F(a, f(a, x))
F(x, f(y, z)) → F(f(x, y), z)

The TRS R consists of the following rules:

f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)

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.a: 0
f: x0
b: 1
F: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

F.1-0(x, f.0-1(y, z)) → F.1-1(f.1-0(x, y), z)
F.0-1(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.1-0(f.1-0(b., a.), x) → F.1-1(b., f.1-0(b., x))
F.0-0(f.0-1(a., b.), x) → F.0-0(a., f.0-0(a., x))
F.1-1(x, f.1-0(y, z)) → F.1-0(f.1-1(x, y), z)
F.0-1(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
F.1-0(x, f.0-0(y, z)) → F.1-0(f.1-0(x, y), z)
F.1-1(f.1-0(b., a.), x) → F.1-1(b., f.1-1(b., x))
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.0-1(f.0-1(a., b.), x) → F.0-0(a., f.0-1(a., x))
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)

The TRS R consists of the following rules:

f.0-1(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.1-0(f.1-0(b., a.), x) → f.1-1(b., f.1-0(b., x))
f.0-1(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.0-1(f.0-1(a., b.), x) → f.0-0(a., f.0-1(a., x))
f.1-0(x, f.0-1(y, z)) → f.1-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.0-1(a., b.), x) → f.0-0(a., f.0-0(a., x))
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)
f.1-0(x, f.0-0(y, z)) → f.1-0(f.1-0(x, y), z)
f.1-1(f.1-0(b., a.), x) → f.1-1(b., f.1-1(b., x))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ SemLabProof
QDP
              ↳ DependencyGraphProof

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

F.1-0(x, f.0-1(y, z)) → F.1-1(f.1-0(x, y), z)
F.0-1(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.1-0(f.1-0(b., a.), x) → F.1-1(b., f.1-0(b., x))
F.0-0(f.0-1(a., b.), x) → F.0-0(a., f.0-0(a., x))
F.1-1(x, f.1-0(y, z)) → F.1-0(f.1-1(x, y), z)
F.0-1(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
F.1-0(x, f.0-0(y, z)) → F.1-0(f.1-0(x, y), z)
F.1-1(f.1-0(b., a.), x) → F.1-1(b., f.1-1(b., x))
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.0-1(f.0-1(a., b.), x) → F.0-0(a., f.0-1(a., x))
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)

The TRS R consists of the following rules:

f.0-1(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.1-0(f.1-0(b., a.), x) → f.1-1(b., f.1-0(b., x))
f.0-1(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.0-1(f.0-1(a., b.), x) → f.0-0(a., f.0-1(a., x))
f.1-0(x, f.0-1(y, z)) → f.1-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.0-1(a., b.), x) → f.0-0(a., f.0-0(a., x))
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)
f.1-0(x, f.0-0(y, z)) → f.1-0(f.1-0(x, y), z)
f.1-1(f.1-0(b., a.), x) → f.1-1(b., f.1-1(b., x))

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
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                    ↳ UsableRulesReductionPairsProof
                  ↳ QDP

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

F.0-1(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.0-0(f.0-1(a., b.), x) → F.0-0(a., f.0-0(a., x))
F.0-1(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.0-1(f.0-1(a., b.), x) → F.0-0(a., f.0-1(a., x))
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)

The TRS R consists of the following rules:

f.0-1(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.1-0(f.1-0(b., a.), x) → f.1-1(b., f.1-0(b., x))
f.0-1(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.0-1(f.0-1(a., b.), x) → f.0-0(a., f.0-1(a., x))
f.1-0(x, f.0-1(y, z)) → f.1-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.0-1(a., b.), x) → f.0-0(a., f.0-0(a., x))
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)
f.1-0(x, f.0-0(y, z)) → f.1-0(f.1-0(x, y), z)
f.1-1(f.1-0(b., a.), x) → f.1-1(b., f.1-1(b., x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

F.0-1(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.0-0(f.0-1(a., b.), x) → F.0-0(a., f.0-0(a., x))
F.0-1(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
F.0-1(f.0-1(a., b.), x) → F.0-0(a., f.0-1(a., x))
The following rules are removed from R:

f.0-1(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-1(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.0-1(f.0-1(a., b.), x) → f.0-0(a., f.0-1(a., x))
f.0-0(f.0-1(a., b.), x) → f.0-0(a., f.0-0(a., x))
Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0-0(x1, x2)) = 1 + x1 + x2   
POL(F.0-1(x1, x2)) = 1 + x1 + x2   
POL(a.) = 0   
POL(b.) = 1   
POL(f.0-0(x1, x2)) = x1 + x2   
POL(f.0-1(x1, x2)) = x1 + x2   
POL(f.1-0(x1, x2)) = 1 + x1 + x2   
POL(f.1-1(x1, x2)) = x1 + x2   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
QDP
                        ↳ DependencyGraphProof
                  ↳ QDP

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

F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)

The TRS R consists of the following rules:

f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)

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
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
                      ↳ QDP
                        ↳ DependencyGraphProof
QDP
                            ↳ QDPOrderProof
                  ↳ QDP

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

F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)

The TRS R consists of the following rules:

f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)

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.


F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(F.0-0(x1, x2)) = x2   
POL(f.0-0(x1, x2)) = 1 + x1 + x2   
POL(f.0-1(x1, x2)) = 0   

The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
                      ↳ QDP
                        ↳ DependencyGraphProof
                          ↳ QDP
                            ↳ QDPOrderProof
QDP
                                ↳ PisEmptyProof
                  ↳ QDP

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

f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)

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
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
QDP
                    ↳ UsableRulesReductionPairsProof

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

F.1-0(x, f.0-1(y, z)) → F.1-1(f.1-0(x, y), z)
F.1-0(f.1-0(b., a.), x) → F.1-1(b., f.1-0(b., x))
F.1-1(x, f.1-0(y, z)) → F.1-0(f.1-1(x, y), z)
F.1-0(x, f.0-0(y, z)) → F.1-0(f.1-0(x, y), z)
F.1-1(f.1-0(b., a.), x) → F.1-1(b., f.1-1(b., x))
F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)

The TRS R consists of the following rules:

f.0-1(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.1-0(f.1-0(b., a.), x) → f.1-1(b., f.1-0(b., x))
f.0-1(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.0-1(f.0-1(a., b.), x) → f.0-0(a., f.0-1(a., x))
f.1-0(x, f.0-1(y, z)) → f.1-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
f.0-0(f.0-1(a., b.), x) → f.0-0(a., f.0-0(a., x))
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)
f.1-0(x, f.0-0(y, z)) → f.1-0(f.1-0(x, y), z)
f.1-1(f.1-0(b., a.), x) → f.1-1(b., f.1-1(b., x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

F.1-0(x, f.0-1(y, z)) → F.1-1(f.1-0(x, y), z)
F.1-0(f.1-0(b., a.), x) → F.1-1(b., f.1-0(b., x))
F.1-0(x, f.0-0(y, z)) → F.1-0(f.1-0(x, y), z)
F.1-1(f.1-0(b., a.), x) → F.1-1(b., f.1-1(b., x))
The following rules are removed from R:

f.1-0(f.1-0(b., a.), x) → f.1-1(b., f.1-0(b., x))
f.1-0(x, f.0-0(y, z)) → f.1-0(f.1-0(x, y), z)
f.1-1(f.1-0(b., a.), x) → f.1-1(b., f.1-1(b., x))
f.1-0(x, f.0-1(y, z)) → f.1-1(f.1-0(x, y), z)
Used ordering: POLO with Polynomial interpretation [25]:

POL(F.1-0(x1, x2)) = 1 + x1 + x2   
POL(F.1-1(x1, x2)) = 1 + x1 + x2   
POL(a.) = 1   
POL(b.) = 0   
POL(f.0-0(x1, x2)) = 1 + x1 + x2   
POL(f.0-1(x1, x2)) = 1 + x1 + x2   
POL(f.1-0(x1, x2)) = x1 + x2   
POL(f.1-1(x1, x2)) = x1 + x2   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
QDP
                        ↳ DependencyGraphProof

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

F.1-1(x, f.1-0(y, z)) → F.1-0(f.1-1(x, y), z)
F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)

The TRS R consists of the following rules:

f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)

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
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
                      ↳ QDP
                        ↳ DependencyGraphProof
QDP
                            ↳ QDPOrderProof

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

F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)

The TRS R consists of the following rules:

f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)

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.


F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(F.1-1(x1, x2)) = x2   
POL(f.1-0(x1, x2)) = 0   
POL(f.1-1(x1, x2)) = 1 + x1 + x2   

The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ SemLabProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ UsableRulesReductionPairsProof
                      ↳ QDP
                        ↳ DependencyGraphProof
                          ↳ QDP
                            ↳ QDPOrderProof
QDP
                                ↳ PisEmptyProof

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

f.1-1(x, f.1-0(y, z)) → f.1-0(f.1-1(x, y), z)
f.1-1(x, f.1-1(y, z)) → f.1-1(f.1-1(x, y), z)

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.