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:
h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> g2(X, Y)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> g2(X, Y)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
H2(X, Z) -> F3(X, s1(X), Z)
G2(X, s1(Y)) -> G2(X, Y)
F3(X, Y, g2(X, Y)) -> H2(0, g2(X, Y))
The TRS R consists of the following rules:
h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> g2(X, Y)
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:
H2(X, Z) -> F3(X, s1(X), Z)
G2(X, s1(Y)) -> G2(X, Y)
F3(X, Y, g2(X, Y)) -> H2(0, g2(X, Y))
The TRS R consists of the following rules:
h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> g2(X, Y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G2(X, s1(Y)) -> G2(X, Y)
The TRS R consists of the following rules:
h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> 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(X, s1(Y)) -> G2(X, Y)
Used argument filtering: G2(x1, x2) = x2
s1(x1) = s1(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:
h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> g2(X, Y)
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:
H2(X, Z) -> F3(X, s1(X), Z)
F3(X, Y, g2(X, Y)) -> H2(0, g2(X, Y))
The TRS R consists of the following rules:
h2(X, Z) -> f3(X, s1(X), Z)
f3(X, Y, g2(X, Y)) -> h2(0, g2(X, Y))
g2(0, Y) -> 0
g2(X, s1(Y)) -> g2(X, Y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.