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(1) -> f1(g1(1))
f1(f1(x)) -> f1(x)
g1(0) -> g1(f1(0))
g1(g1(x)) -> g1(x)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f1(1) -> f1(g1(1))
f1(f1(x)) -> f1(x)
g1(0) -> g1(f1(0))
g1(g1(x)) -> g1(x)

Q is empty.

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

F1(1) -> F1(g1(1))
G1(0) -> G1(f1(0))
F1(1) -> G1(1)
G1(0) -> F1(0)

The TRS R consists of the following rules:

f1(1) -> f1(g1(1))
f1(f1(x)) -> f1(x)
g1(0) -> g1(f1(0))
g1(g1(x)) -> 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(1) -> F1(g1(1))
G1(0) -> G1(f1(0))
F1(1) -> G1(1)
G1(0) -> F1(0)

The TRS R consists of the following rules:

f1(1) -> f1(g1(1))
f1(f1(x)) -> f1(x)
g1(0) -> g1(f1(0))
g1(g1(x)) -> g1(x)

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