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), g(y)) → f(p(f(g(x), s(y))), g(s(p(x))))
p(0) → g(0)
g(s(p(x))) → p(x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(x), g(y)) → f(p(f(g(x), s(y))), g(s(p(x))))
p(0) → g(0)
g(s(p(x))) → p(x)
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F(g(x), g(y)) → P(f(g(x), s(y)))
F(g(x), g(y)) → F(p(f(g(x), s(y))), g(s(p(x))))
F(g(x), g(y)) → F(g(x), s(y))
P(0) → G(0)
F(g(x), g(y)) → G(s(p(x)))
F(g(x), g(y)) → P(x)
The TRS R consists of the following rules:
f(g(x), g(y)) → f(p(f(g(x), s(y))), g(s(p(x))))
p(0) → g(0)
g(s(p(x))) → p(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
F(g(x), g(y)) → P(f(g(x), s(y)))
F(g(x), g(y)) → F(p(f(g(x), s(y))), g(s(p(x))))
F(g(x), g(y)) → F(g(x), s(y))
P(0) → G(0)
F(g(x), g(y)) → G(s(p(x)))
F(g(x), g(y)) → P(x)
The TRS R consists of the following rules:
f(g(x), g(y)) → f(p(f(g(x), s(y))), g(s(p(x))))
p(0) → g(0)
g(s(p(x))) → p(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(g(x), g(y)) → F(p(f(g(x), s(y))), g(s(p(x))))
F(g(x), g(y)) → P(f(g(x), s(y)))
F(g(x), g(y)) → F(g(x), s(y))
P(0) → G(0)
F(g(x), g(y)) → G(s(p(x)))
F(g(x), g(y)) → P(x)
The TRS R consists of the following rules:
f(g(x), g(y)) → f(p(f(g(x), s(y))), g(s(p(x))))
p(0) → g(0)
g(s(p(x))) → p(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 6 less nodes.