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 strictly oriented and are deleted.


F1(g1(f1(X))) -> F1(g1(X))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F1(x1)  =  F1(x1)
g1(x1)  =  x1
f1(x1)  =  f1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


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.