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(g1(X)) -> g1(f1(f1(X)))
f1(h1(X)) -> h1(g1(X))

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

f1(g1(X)) -> g1(f1(f1(X)))
f1(h1(X)) -> h1(g1(X))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

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

f1(g1(X)) -> g1(f1(f1(X)))
f1(h1(X)) -> h1(g1(X))

The set Q consists of the following terms:

f1(g1(x0))
f1(h1(x0))


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

The TRS R consists of the following rules:

f1(g1(X)) -> g1(f1(f1(X)))
f1(h1(X)) -> h1(g1(X))

The set Q consists of the following terms:

f1(g1(x0))
f1(h1(x0))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

f1(g1(X)) -> g1(f1(f1(X)))
f1(h1(X)) -> h1(g1(X))

The set Q consists of the following terms:

f1(g1(x0))
f1(h1(x0))

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(X)) -> F1(f1(X))
F1(g1(X)) -> F1(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)  =  g1(x1)
f1(x1)  =  x1
h1(x1)  =  h

Lexicographic Path Order [19].
Precedence:
[F1, g1, h]


The following usable rules [14] were oriented:

f1(g1(X)) -> g1(f1(f1(X)))
f1(h1(X)) -> h1(g1(X))



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
QDP
              ↳ PisEmptyProof

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

f1(g1(X)) -> g1(f1(f1(X)))
f1(h1(X)) -> h1(g1(X))

The set Q consists of the following terms:

f1(g1(x0))
f1(h1(x0))

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