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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 3 less nodes.