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

Q is empty.


↳ QTRS
  ↳ AAECC Innermost

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

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

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

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

The signature Sigma is {f}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

f(x0, a(b(x1)))
f(x0, b(a(x1)))
f(a(x0), x1)
f(b(x0), x1)


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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, a(b(x1)))
f(x0, b(a(x1)))
f(a(x0), x1)
f(b(x0), x1)

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, a(b(x1)))
f(x0, b(a(x1)))
f(a(x0), x1)
f(b(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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

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

R is empty.
The set Q consists of the following terms:

f(x0, a(b(x1)))
f(x0, b(a(x1)))
f(a(x0), x1)
f(b(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(x0, a(b(x1)))
f(x0, b(a(x1)))
f(a(x0), x1)
f(b(x0), x1)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof

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

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

R is empty.
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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof

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

a(b(x)) → a(a(x))
b(a(x)) → b(b(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:

B(a(x)) → B(b(x))
A(b(x)) → A(a(x))
B(a(x)) → B(x)
A(b(x)) → A(x)

The TRS R consists of the following rules:

a(b(x)) → a(a(x))
b(a(x)) → b(b(x))

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof

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

B(a(x)) → B(b(x))
A(b(x)) → A(a(x))
B(a(x)) → B(x)
A(b(x)) → A(x)

The TRS R consists of the following rules:

a(b(x)) → a(a(x))
b(a(x)) → b(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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                                ↳ UsableRulesProof
                              ↳ QDP

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

B(a(x)) → B(b(x))
B(a(x)) → B(x)

The TRS R consists of the following rules:

a(b(x)) → a(a(x))
b(a(x)) → b(b(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                              ↳ QDP

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

B(a(x)) → B(b(x))
B(a(x)) → B(x)

The TRS R consists of the following rules:

b(a(x)) → b(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 1 SCC with 1 less node.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ MNOCProof
                              ↳ QDP

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

B(a(x)) → B(x)

The TRS R consists of the following rules:

b(a(x)) → b(b(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ MNOCProof
                                          ↳ QDP
                                            ↳ UsableRulesProof
                              ↳ QDP

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

B(a(x)) → B(x)

The TRS R consists of the following rules:

b(a(x)) → b(b(x))

The set Q consists of the following terms:

b(a(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ MNOCProof
                                          ↳ QDP
                                            ↳ UsableRulesProof
                                              ↳ QDP
                                                ↳ QReductionProof
                              ↳ QDP

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

B(a(x)) → B(x)

R is empty.
The set Q consists of the following terms:

b(a(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

b(a(x0))



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ MNOCProof
                                          ↳ QDP
                                            ↳ UsableRulesProof
                                              ↳ QDP
                                                ↳ QReductionProof
                                                  ↳ QDP
                                                    ↳ QDPSizeChangeProof
                              ↳ QDP

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

B(a(x)) → B(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                              ↳ QDP
                                ↳ UsableRulesProof

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

A(b(x)) → A(a(x))
A(b(x)) → A(x)

The TRS R consists of the following rules:

a(b(x)) → a(a(x))
b(a(x)) → b(b(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ DependencyGraphProof

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

A(b(x)) → A(a(x))
A(b(x)) → A(x)

The TRS R consists of the following rules:

a(b(x)) → a(a(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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ MNOCProof

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

A(b(x)) → A(x)

The TRS R consists of the following rules:

a(b(x)) → a(a(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ MNOCProof
                                          ↳ QDP
                                            ↳ UsableRulesProof

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

A(b(x)) → A(x)

The TRS R consists of the following rules:

a(b(x)) → a(a(x))

The set Q consists of the following terms:

a(b(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ MNOCProof
                                          ↳ QDP
                                            ↳ UsableRulesProof
                                              ↳ QDP
                                                ↳ QReductionProof

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

A(b(x)) → A(x)

R is empty.
The set Q consists of the following terms:

a(b(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

a(b(x0))



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ AND
                              ↳ QDP
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ MNOCProof
                                          ↳ QDP
                                            ↳ UsableRulesProof
                                              ↳ QDP
                                                ↳ QReductionProof
                                                  ↳ QDP
                                                    ↳ QDPSizeChangeProof

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

A(b(x)) → A(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: