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:

f1(f1(X)) -> f1(g1(f1(g1(f1(X)))))
f1(g1(f1(X))) -> f1(g1(X))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f1(f1(X)) -> f1(g1(f1(g1(f1(X)))))
f1(g1(f1(X))) -> f1(g1(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:

F1(g1(f1(X))) -> F1(g1(X))
F1(f1(X)) -> F1(g1(f1(g1(f1(X)))))
F1(f1(X)) -> F1(g1(f1(X)))

The TRS R consists of the following rules:

f1(f1(X)) -> f1(g1(f1(g1(f1(X)))))
f1(g1(f1(X))) -> f1(g1(X))

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:

F1(g1(f1(X))) -> F1(g1(X))
F1(f1(X)) -> F1(g1(f1(g1(f1(X)))))
F1(f1(X)) -> F1(g1(f1(X)))

The TRS R consists of the following rules:

f1(f1(X)) -> f1(g1(f1(g1(f1(X)))))
f1(g1(f1(X))) -> f1(g1(X))

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 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

F1(g1(f1(X))) -> F1(g1(X))

The TRS R consists of the following rules:

f1(f1(X)) -> f1(g1(f1(g1(f1(X)))))
f1(g1(f1(X))) -> f1(g1(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.


F1(g1(f1(X))) -> F1(g1(X))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F1(x1)) = 2·x1   
POL(f1(x1)) = 2 + 2·x1 + 2·x12   
POL(g1(x1)) = x1   

The following usable rules [14] were oriented: none



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

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

f1(f1(X)) -> f1(g1(f1(g1(f1(X)))))
f1(g1(f1(X))) -> f1(g1(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.