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:
f1(c3(a, z, x)) -> b2(a, z)
b2(x, b2(z, y)) -> f1(b2(f1(f1(z)), c3(x, z, y)))
b2(y, z) -> z
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(c3(a, z, x)) -> b2(a, z)
b2(x, b2(z, y)) -> f1(b2(f1(f1(z)), c3(x, z, y)))
b2(y, z) -> z
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F1(c3(a, z, x)) -> B2(a, z)
B2(x, b2(z, y)) -> F1(b2(f1(f1(z)), c3(x, z, y)))
B2(x, b2(z, y)) -> F1(f1(z))
B2(x, b2(z, y)) -> F1(z)
B2(x, b2(z, y)) -> B2(f1(f1(z)), c3(x, z, y))
The TRS R consists of the following rules:
f1(c3(a, z, x)) -> b2(a, z)
b2(x, b2(z, y)) -> f1(b2(f1(f1(z)), c3(x, z, y)))
b2(y, z) -> z
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:
F1(c3(a, z, x)) -> B2(a, z)
B2(x, b2(z, y)) -> F1(b2(f1(f1(z)), c3(x, z, y)))
B2(x, b2(z, y)) -> F1(f1(z))
B2(x, b2(z, y)) -> F1(z)
B2(x, b2(z, y)) -> B2(f1(f1(z)), c3(x, z, y))
The TRS R consists of the following rules:
f1(c3(a, z, x)) -> b2(a, z)
b2(x, b2(z, y)) -> f1(b2(f1(f1(z)), c3(x, z, y)))
b2(y, z) -> z
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F1(c3(a, z, x)) -> B2(a, z)
B2(x, b2(z, y)) -> F1(b2(f1(f1(z)), c3(x, z, y)))
B2(x, b2(z, y)) -> F1(f1(z))
B2(x, b2(z, y)) -> F1(z)
The TRS R consists of the following rules:
f1(c3(a, z, x)) -> b2(a, z)
b2(x, b2(z, y)) -> f1(b2(f1(f1(z)), c3(x, z, y)))
b2(y, z) -> z
Q is empty.
We have to consider all minimal (P,Q,R)-chains.