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:

p1(f1(f1(x))) -> q1(f1(g1(x)))
p1(g1(g1(x))) -> q1(g1(f1(x)))
q1(f1(f1(x))) -> p1(f1(g1(x)))
q1(g1(g1(x))) -> p1(g1(f1(x)))

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

p1(f1(f1(x))) -> q1(f1(g1(x)))
p1(g1(g1(x))) -> q1(g1(f1(x)))
q1(f1(f1(x))) -> p1(f1(g1(x)))
q1(g1(g1(x))) -> p1(g1(f1(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:

p1(f1(f1(x))) -> q1(f1(g1(x)))
p1(g1(g1(x))) -> q1(g1(f1(x)))
q1(f1(f1(x))) -> p1(f1(g1(x)))
q1(g1(g1(x))) -> p1(g1(f1(x)))

The set Q consists of the following terms:

p1(f1(f1(x0)))
p1(g1(g1(x0)))
q1(f1(f1(x0)))
q1(g1(g1(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:

Q1(g1(g1(x))) -> P1(g1(f1(x)))
P1(f1(f1(x))) -> Q1(f1(g1(x)))
P1(g1(g1(x))) -> Q1(g1(f1(x)))
Q1(f1(f1(x))) -> P1(f1(g1(x)))

The TRS R consists of the following rules:

p1(f1(f1(x))) -> q1(f1(g1(x)))
p1(g1(g1(x))) -> q1(g1(f1(x)))
q1(f1(f1(x))) -> p1(f1(g1(x)))
q1(g1(g1(x))) -> p1(g1(f1(x)))

The set Q consists of the following terms:

p1(f1(f1(x0)))
p1(g1(g1(x0)))
q1(f1(f1(x0)))
q1(g1(g1(x0)))

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

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

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

Q1(g1(g1(x))) -> P1(g1(f1(x)))
P1(f1(f1(x))) -> Q1(f1(g1(x)))
P1(g1(g1(x))) -> Q1(g1(f1(x)))
Q1(f1(f1(x))) -> P1(f1(g1(x)))

The TRS R consists of the following rules:

p1(f1(f1(x))) -> q1(f1(g1(x)))
p1(g1(g1(x))) -> q1(g1(f1(x)))
q1(f1(f1(x))) -> p1(f1(g1(x)))
q1(g1(g1(x))) -> p1(g1(f1(x)))

The set Q consists of the following terms:

p1(f1(f1(x0)))
p1(g1(g1(x0)))
q1(f1(f1(x0)))
q1(g1(g1(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 4 less nodes.