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(f(a, a), a)) → f(f(a, f(a, a)), x)

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

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

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

f(x0, f(f(a, a), a))


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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(f(a, a), a))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(f(a, a), a))

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(f(a, a), a))

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ RuleRemovalProof

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

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

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

f(x0, f(f(a, a), a))

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


Used ordering: POLO with Polynomial interpretation [25]:

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ RuleRemovalProof
                    ↳ QDP
                      ↳ PisEmptyProof

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

f(x0, f(f(a, a), a))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.