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:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(x)
The set Q consists of the following terms:
f(g(x0), x0, x1)
g(g(x0))
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(x)
The set Q consists of the following terms:
f(g(x0), x0, x1)
g(g(x0))
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F(g(x), x, y) → F(y, y, g(y))
F(g(x), x, y) → G(y)
The TRS R consists of the following rules:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(x)
The set Q consists of the following terms:
f(g(x0), x0, x1)
g(g(x0))
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:
F(g(x), x, y) → F(y, y, g(y))
F(g(x), x, y) → G(y)
The TRS R consists of the following rules:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(x)
The set Q consists of the following terms:
f(g(x0), x0, x1)
g(g(x0))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.