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) -> g2(x, y)
g2(h1(x), y) -> h1(f2(x, y))
g2(h1(x), y) -> h1(g2(x, y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(x, y) -> g2(x, y)
g2(h1(x), y) -> h1(f2(x, y))
g2(h1(x), y) -> h1(g2(x, y))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
G2(h1(x), y) -> G2(x, y)
F2(x, y) -> G2(x, y)
G2(h1(x), y) -> F2(x, y)
The TRS R consists of the following rules:
f2(x, y) -> g2(x, y)
g2(h1(x), y) -> h1(f2(x, y))
g2(h1(x), y) -> h1(g2(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
G2(h1(x), y) -> G2(x, y)
F2(x, y) -> G2(x, y)
G2(h1(x), y) -> F2(x, y)
The TRS R consists of the following rules:
f2(x, y) -> g2(x, y)
g2(h1(x), y) -> h1(f2(x, y))
g2(h1(x), y) -> h1(g2(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
G2(h1(x), y) -> G2(x, y)
G2(h1(x), y) -> F2(x, y)
Used argument filtering: G2(x1, x2) = x1
h1(x1) = h1(x1)
F2(x1, x2) = x1
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F2(x, y) -> G2(x, y)
The TRS R consists of the following rules:
f2(x, y) -> g2(x, y)
g2(h1(x), y) -> h1(f2(x, y))
g2(h1(x), y) -> h1(g2(x, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.