Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(x)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F4(0, 1, g2(x, y), z) -> H1(x)
H1(g2(x, y)) -> H1(x)
F4(0, 1, g2(x, y), z) -> F4(g2(x, y), g2(x, y), g2(x, y), h1(x))
The TRS R consists of the following rules:
f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(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:
F4(0, 1, g2(x, y), z) -> H1(x)
H1(g2(x, y)) -> H1(x)
F4(0, 1, g2(x, y), z) -> F4(g2(x, y), g2(x, y), g2(x, y), h1(x))
The TRS R consists of the following rules:
f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
H1(g2(x, y)) -> H1(x)
The TRS R consists of the following rules:
f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(x)
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.
H1(g2(x, y)) -> H1(x)
Used argument filtering: H1(x1) = x1
g2(x1, x2) = g1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F4(0, 1, g2(x, y), z) -> F4(g2(x, y), g2(x, y), g2(x, y), h1(x))
The TRS R consists of the following rules:
f4(0, 1, g2(x, y), z) -> f4(g2(x, y), g2(x, y), g2(x, y), h1(x))
g2(0, 1) -> 0
g2(0, 1) -> 1
h1(g2(x, y)) -> h1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.