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:
f2(a, f2(a, x)) -> f2(a, f2(f2(a, a), f2(a, x)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(a, f2(a, x)) -> f2(a, f2(f2(a, a), f2(a, x)))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F2(a, f2(a, x)) -> F2(a, f2(f2(a, a), f2(a, x)))
F2(a, f2(a, x)) -> F2(a, a)
F2(a, f2(a, x)) -> F2(f2(a, a), f2(a, x))
The TRS R consists of the following rules:
f2(a, f2(a, x)) -> f2(a, f2(f2(a, a), f2(a, 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:
F2(a, f2(a, x)) -> F2(a, f2(f2(a, a), f2(a, x)))
F2(a, f2(a, x)) -> F2(a, a)
F2(a, f2(a, x)) -> F2(f2(a, a), f2(a, x))
The TRS R consists of the following rules:
f2(a, f2(a, x)) -> f2(a, f2(f2(a, a), f2(a, 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.