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

The set Q consists of the following terms:

f(x0, f(a, x1))



↳ QTRS
  ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

f(x0, f(a, 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, f(a, y)) → F(a, a)
F(x, f(a, y)) → F(a, f(f(f(a, a), y), x))
F(x, f(a, y)) → F(f(f(a, a), y), x)
F(x, f(a, y)) → F(f(a, a), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(a, x1))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(a, x1))

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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(a, x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule F(x, f(a, y)) → F(a, f(f(f(a, a), y), x)) at position [1] we obtained the following new rules:

F(f(a, x1), f(a, y1)) → F(a, f(a, f(f(f(a, a), x1), f(f(a, a), y1))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof

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

F(f(a, x1), f(a, y1)) → F(a, f(a, f(f(f(a, a), x1), f(f(a, a), y1))))
F(x, f(a, y)) → F(f(f(a, a), y), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(a, x1))

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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ UsableRulesProof

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

F(x, f(a, y)) → F(f(f(a, a), y), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(a, 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ Instantiation

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

F(x, f(a, y)) → F(f(f(a, a), y), x)

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

f(x0, f(a, x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(x, f(a, y)) → F(f(f(a, a), y), x) we obtained the following new rules:

F(f(f(a, a), z1), f(a, x1)) → F(f(f(a, a), x1), f(f(a, a), z1))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ UsableRulesProof
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ DependencyGraphProof

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

F(f(f(a, a), z1), f(a, x1)) → F(f(f(a, a), x1), f(f(a, a), z1))

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

f(x0, f(a, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.