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:
f5(x, y, w, w, a) -> g14(x, x, y, w)
f5(x, y, w, a, a) -> g14(y, x, x, w)
f5(x, y, a, a, w) -> g24(x, y, y, w)
f5(x, y, a, w, w) -> g24(y, y, x, w)
g14(x, x, y, a) -> h2(x, y)
g14(y, x, x, a) -> h2(x, y)
g24(x, y, y, a) -> h2(x, y)
g24(y, y, x, a) -> h2(x, y)
h2(x, x) -> x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f5(x, y, w, w, a) -> g14(x, x, y, w)
f5(x, y, w, a, a) -> g14(y, x, x, w)
f5(x, y, a, a, w) -> g24(x, y, y, w)
f5(x, y, a, w, w) -> g24(y, y, x, w)
g14(x, x, y, a) -> h2(x, y)
g14(y, x, x, a) -> h2(x, y)
g24(x, y, y, a) -> h2(x, y)
g24(y, y, x, a) -> h2(x, y)
h2(x, x) -> x
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F5(x, y, a, a, w) -> G24(x, y, y, w)
F5(x, y, w, w, a) -> G14(x, x, y, w)
G14(y, x, x, a) -> H2(x, y)
G14(x, x, y, a) -> H2(x, y)
G24(y, y, x, a) -> H2(x, y)
F5(x, y, w, a, a) -> G14(y, x, x, w)
F5(x, y, a, w, w) -> G24(y, y, x, w)
G24(x, y, y, a) -> H2(x, y)
The TRS R consists of the following rules:
f5(x, y, w, w, a) -> g14(x, x, y, w)
f5(x, y, w, a, a) -> g14(y, x, x, w)
f5(x, y, a, a, w) -> g24(x, y, y, w)
f5(x, y, a, w, w) -> g24(y, y, x, w)
g14(x, x, y, a) -> h2(x, y)
g14(y, x, x, a) -> h2(x, y)
g24(x, y, y, a) -> h2(x, y)
g24(y, y, x, a) -> h2(x, y)
h2(x, x) -> 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:
F5(x, y, a, a, w) -> G24(x, y, y, w)
F5(x, y, w, w, a) -> G14(x, x, y, w)
G14(y, x, x, a) -> H2(x, y)
G14(x, x, y, a) -> H2(x, y)
G24(y, y, x, a) -> H2(x, y)
F5(x, y, w, a, a) -> G14(y, x, x, w)
F5(x, y, a, w, w) -> G24(y, y, x, w)
G24(x, y, y, a) -> H2(x, y)
The TRS R consists of the following rules:
f5(x, y, w, w, a) -> g14(x, x, y, w)
f5(x, y, w, a, a) -> g14(y, x, x, w)
f5(x, y, a, a, w) -> g24(x, y, y, w)
f5(x, y, a, w, w) -> g24(y, y, x, w)
g14(x, x, y, a) -> h2(x, y)
g14(y, x, x, a) -> h2(x, y)
g24(x, y, y, a) -> h2(x, y)
g24(y, y, x, a) -> h2(x, y)
h2(x, x) -> x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 8 less nodes.