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)) -> g1(f1(x))
g1(g1(x)) -> f1(x)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f1(f1(x)) -> g1(f1(x))
g1(g1(x)) -> f1(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(f1(x)) -> G1(f1(x))
G1(g1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(f1(x)) -> g1(f1(x))
g1(g1(x)) -> f1(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:

F1(f1(x)) -> G1(f1(x))
G1(g1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(f1(x)) -> g1(f1(x))
g1(g1(x)) -> f1(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.


G1(g1(x)) -> F1(x)
The remaining pairs can at least by weakly be oriented.

F1(f1(x)) -> G1(f1(x))
Used ordering: Combined order from the following AFS and order.
F1(x1)  =  x1
f1(x1)  =  f1(x1)
G1(x1)  =  x1
g1(x1)  =  g1(x1)

Lexicographic Path Order [19].
Precedence:
[f1, g1]


The following usable rules [14] were oriented:

g1(g1(x)) -> f1(x)
f1(f1(x)) -> g1(f1(x))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ DependencyGraphProof

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

F1(f1(x)) -> G1(f1(x))

The TRS R consists of the following rules:

f1(f1(x)) -> g1(f1(x))
g1(g1(x)) -> f1(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.