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:
p(f(f(x))) → q(f(g(x)))
p(g(g(x))) → q(g(f(x)))
q(f(f(x))) → p(f(g(x)))
q(g(g(x))) → p(g(f(x)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
p(f(f(x))) → q(f(g(x)))
p(g(g(x))) → q(g(f(x)))
q(f(f(x))) → p(f(g(x)))
q(g(g(x))) → p(g(f(x)))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
Q(g(g(x))) → P(g(f(x)))
P(g(g(x))) → Q(g(f(x)))
Q(f(f(x))) → P(f(g(x)))
P(f(f(x))) → Q(f(g(x)))
The TRS R consists of the following rules:
p(f(f(x))) → q(f(g(x)))
p(g(g(x))) → q(g(f(x)))
q(f(f(x))) → p(f(g(x)))
q(g(g(x))) → p(g(f(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:
Q(g(g(x))) → P(g(f(x)))
P(g(g(x))) → Q(g(f(x)))
Q(f(f(x))) → P(f(g(x)))
P(f(f(x))) → Q(f(g(x)))
The TRS R consists of the following rules:
p(f(f(x))) → q(f(g(x)))
p(g(g(x))) → q(g(f(x)))
q(f(f(x))) → p(f(g(x)))
q(g(g(x))) → p(g(f(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 4 less nodes.