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(a1(b1(f1(X))))
f1(a1(g1(X))) -> b1(X)
b1(X) -> a1(X)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f1(f1(X)) -> f1(a1(b1(f1(X))))
f1(a1(g1(X))) -> b1(X)
b1(X) -> a1(X)

Q is empty.

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

F1(f1(X)) -> B1(f1(X))
F1(a1(g1(X))) -> B1(X)
F1(f1(X)) -> F1(a1(b1(f1(X))))

The TRS R consists of the following rules:

f1(f1(X)) -> f1(a1(b1(f1(X))))
f1(a1(g1(X))) -> b1(X)
b1(X) -> a1(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(f1(X)) -> B1(f1(X))
F1(a1(g1(X))) -> B1(X)
F1(f1(X)) -> F1(a1(b1(f1(X))))

The TRS R consists of the following rules:

f1(f1(X)) -> f1(a1(b1(f1(X))))
f1(a1(g1(X))) -> b1(X)
b1(X) -> a1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 3 less nodes.