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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
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(f(y, z), f(x, f(a, x))) → F(f(a, z), f(x, a))
F(f(y, z), f(x, f(a, x))) → F(f(f(a, z), f(x, a)), f(a, y))
F(f(y, z), f(x, f(a, x))) → F(x, a)
F(f(y, z), f(x, f(a, x))) → F(a, z)
F(f(y, z), f(x, f(a, x))) → F(a, y)

The TRS R consists of the following rules:

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

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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

f(f(y, z), f(x, f(a, x))) → f(f(f(a, z), f(x, a)), f(a, 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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ Instantiation

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

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

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

F(f(f(a, z1), f(z2, a)), f(a, f(a, a))) → F(f(f(a, f(z2, a)), f(a, a)), f(a, f(a, z1)))



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

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

F(f(f(a, z1), f(z2, a)), f(a, f(a, a))) → F(f(f(a, f(z2, a)), f(a, a)), f(a, f(a, z1)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(f(f(a, z1), f(z2, a)), f(a, f(a, a))) → F(f(f(a, f(z2, a)), f(a, a)), f(a, f(a, z1))) we obtained the following new rules:

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



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

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

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

R is empty.
Q is empty.
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.