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(g(x)) → g(g(f(x)))
f(g(x)) → g(g(g(x)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(g(x)) → g(g(f(x)))
f(g(x)) → g(g(g(x)))

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(g(x)) → F(x)

The TRS R consists of the following rules:

f(g(x)) → g(g(f(x)))
f(g(x)) → g(g(g(x)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

F(g(x)) → F(x)

The TRS R consists of the following rules:

f(g(x)) → g(g(f(x)))
f(g(x)) → g(g(g(x)))

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

Recursive path order with status [2].
Precedence:
g1 > F1

Status:
g1: multiset
F1: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ PisEmptyProof

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

f(g(x)) → g(g(f(x)))
f(g(x)) → g(g(g(x)))

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.