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:
f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> y
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F3(s1(x), y, z) -> F3(x, s1(c1(y)), c1(z))
F3(x, c1(x), c1(y)) -> F3(y, y, f3(y, x, y))
F3(x, c1(x), c1(y)) -> F3(y, x, y)
The TRS R consists of the following rules:
f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> 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:
F3(s1(x), y, z) -> F3(x, s1(c1(y)), c1(z))
F3(x, c1(x), c1(y)) -> F3(y, y, f3(y, x, y))
F3(x, c1(x), c1(y)) -> F3(y, x, y)
The TRS R consists of the following rules:
f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> 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:
F3(s1(x), y, z) -> F3(x, s1(c1(y)), c1(z))
The TRS R consists of the following rules:
f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> 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.
F3(s1(x), y, z) -> F3(x, s1(c1(y)), c1(z))
Used argument filtering: F3(x1, x2, x3) = x1
s1(x1) = s1(x1)
Used ordering: Quasi 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:
f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> 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:
F3(x, c1(x), c1(y)) -> F3(y, y, f3(y, x, y))
F3(x, c1(x), c1(y)) -> F3(y, x, y)
The TRS R consists of the following rules:
f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.