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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

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

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

F(a, f(f(a, x), a)) → F(a, f(a, x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F(a, f(f(a, x), a)) → F(a, f(a, x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  F(x1, x2)
a  =  a
f(x1, x2)  =  f(x1, x2)

Recursive Path Order [2].
Precedence:
a > [F2, f2]


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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