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(s1(a), s1(b), x) -> f3(x, x, x)
g1(f3(s1(x), s1(y), z)) -> g1(f3(x, y, z))
cons2(x, y) -> x
cons2(x, y) -> y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f3(s1(a), s1(b), x) -> f3(x, x, x)
g1(f3(s1(x), s1(y), z)) -> g1(f3(x, y, z))
cons2(x, y) -> x
cons2(x, y) -> y
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F3(s1(a), s1(b), x) -> F3(x, x, x)
G1(f3(s1(x), s1(y), z)) -> F3(x, y, z)
G1(f3(s1(x), s1(y), z)) -> G1(f3(x, y, z))
The TRS R consists of the following rules:
f3(s1(a), s1(b), x) -> f3(x, x, x)
g1(f3(s1(x), s1(y), z)) -> g1(f3(x, y, z))
cons2(x, y) -> x
cons2(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(a), s1(b), x) -> F3(x, x, x)
G1(f3(s1(x), s1(y), z)) -> F3(x, y, z)
G1(f3(s1(x), s1(y), z)) -> G1(f3(x, y, z))
The TRS R consists of the following rules:
f3(s1(a), s1(b), x) -> f3(x, x, x)
g1(f3(s1(x), s1(y), z)) -> g1(f3(x, y, z))
cons2(x, y) -> x
cons2(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 with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F3(s1(a), s1(b), x) -> F3(x, x, x)
The TRS R consists of the following rules:
f3(s1(a), s1(b), x) -> f3(x, x, x)
g1(f3(s1(x), s1(y), z)) -> g1(f3(x, y, z))
cons2(x, y) -> x
cons2(x, y) -> y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G1(f3(s1(x), s1(y), z)) -> G1(f3(x, y, z))
The TRS R consists of the following rules:
f3(s1(a), s1(b), x) -> f3(x, x, x)
g1(f3(s1(x), s1(y), z)) -> g1(f3(x, y, z))
cons2(x, y) -> x
cons2(x, y) -> y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.