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:
c(c(z, y, a), a, a) → b(z, y)
f(c(x, y, z)) → c(z, f(b(y, z)), a)
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
c(c(z, y, a), a, a) → b(z, y)
f(c(x, y, z)) → c(z, f(b(y, z)), a)
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, x)
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:
B(z, b(c(a, y, a), f(f(x)))) → C(y, a, z)
F(c(x, y, z)) → B(y, z)
F(c(x, y, z)) → C(z, f(b(y, z)), a)
C(c(z, y, a), a, a) → B(z, y)
B(z, b(c(a, y, a), f(f(x)))) → C(c(y, a, z), z, x)
F(c(x, y, z)) → F(b(y, z))
The TRS R consists of the following rules:
c(c(z, y, a), a, a) → b(z, y)
f(c(x, y, z)) → c(z, f(b(y, z)), a)
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
B(z, b(c(a, y, a), f(f(x)))) → C(y, a, z)
F(c(x, y, z)) → B(y, z)
F(c(x, y, z)) → C(z, f(b(y, z)), a)
C(c(z, y, a), a, a) → B(z, y)
B(z, b(c(a, y, a), f(f(x)))) → C(c(y, a, z), z, x)
F(c(x, y, z)) → F(b(y, z))
The TRS R consists of the following rules:
c(c(z, y, a), a, a) → b(z, y)
f(c(x, y, z)) → c(z, f(b(y, z)), a)
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
B(z, b(c(a, y, a), f(f(x)))) → C(y, a, z)
F(c(x, y, z)) → B(y, z)
F(c(x, y, z)) → C(z, f(b(y, z)), a)
C(c(z, y, a), a, a) → B(z, y)
F(c(x, y, z)) → F(b(y, z))
B(z, b(c(a, y, a), f(f(x)))) → C(c(y, a, z), z, x)
The TRS R consists of the following rules:
c(c(z, y, a), a, a) → b(z, y)
f(c(x, y, z)) → c(z, f(b(y, z)), a)
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(z, b(c(a, y, a), f(f(x)))) → C(y, a, z)
C(c(z, y, a), a, a) → B(z, y)
B(z, b(c(a, y, a), f(f(x)))) → C(c(y, a, z), z, x)
The TRS R consists of the following rules:
c(c(z, y, a), a, a) → b(z, y)
f(c(x, y, z)) → c(z, f(b(y, z)), a)
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(c(x, y, z)) → F(b(y, z))
The TRS R consists of the following rules:
c(c(z, y, a), a, a) → b(z, y)
f(c(x, y, z)) → c(z, f(b(y, z)), a)
b(z, b(c(a, y, a), f(f(x)))) → c(c(y, a, z), z, x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.