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(x, y) -> g13(x, x, y)
f2(x, y) -> g13(y, x, x)
f2(x, y) -> g23(x, y, y)
f2(x, y) -> g23(y, y, x)
g13(x, x, y) -> h2(x, y)
g13(y, x, x) -> h2(x, y)
g23(x, y, y) -> h2(x, y)
g23(y, y, x) -> h2(x, y)
h2(x, x) -> x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(x, y) -> g13(x, x, y)
f2(x, y) -> g13(y, x, x)
f2(x, y) -> g23(x, y, y)
f2(x, y) -> g23(y, y, x)
g13(x, x, y) -> h2(x, y)
g13(y, x, x) -> h2(x, y)
g23(x, y, y) -> h2(x, y)
g23(y, y, x) -> h2(x, y)
h2(x, x) -> x
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
G23(x, y, y) -> H2(x, y)
F2(x, y) -> G13(y, x, x)
F2(x, y) -> G13(x, x, y)
G13(x, x, y) -> H2(x, y)
G13(y, x, x) -> H2(x, y)
F2(x, y) -> G23(y, y, x)
F2(x, y) -> G23(x, y, y)
G23(y, y, x) -> H2(x, y)
The TRS R consists of the following rules:
f2(x, y) -> g13(x, x, y)
f2(x, y) -> g13(y, x, x)
f2(x, y) -> g23(x, y, y)
f2(x, y) -> g23(y, y, x)
g13(x, x, y) -> h2(x, y)
g13(y, x, x) -> h2(x, y)
g23(x, y, y) -> h2(x, y)
g23(y, y, x) -> 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:
G23(x, y, y) -> H2(x, y)
F2(x, y) -> G13(y, x, x)
F2(x, y) -> G13(x, x, y)
G13(x, x, y) -> H2(x, y)
G13(y, x, x) -> H2(x, y)
F2(x, y) -> G23(y, y, x)
F2(x, y) -> G23(x, y, y)
G23(y, y, x) -> H2(x, y)
The TRS R consists of the following rules:
f2(x, y) -> g13(x, x, y)
f2(x, y) -> g13(y, x, x)
f2(x, y) -> g23(x, y, y)
f2(x, y) -> g23(y, y, x)
g13(x, x, y) -> h2(x, y)
g13(y, x, x) -> h2(x, y)
g23(x, y, y) -> h2(x, y)
g23(y, y, x) -> 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.