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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesProof
        ↳ QDP
          ↳ QDPToSRSProof

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

F(a(x), y) → F(x, a(y))
F(x, a(b(y))) → F(a(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ DependencyPairsProof

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

a(b(x)) → b(a(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:

A(b(x)) → A(x)

The TRS R consists of the following rules:

a(b(x)) → b(a(x))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ 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)) → b(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ 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)) → b(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesProof
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ 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: