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:
b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(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:
F(f(c(a, y, z))) → B(y, b(z, z))
C(c(z, x, a), a, y) → F(f(c(y, a, f(c(z, y, x)))))
C(c(z, x, a), a, y) → F(c(z, y, x))
F(f(c(a, y, z))) → B(z, z)
C(c(z, x, a), a, y) → F(c(y, a, f(c(z, y, x))))
C(c(z, x, a), a, y) → C(z, y, x)
C(c(z, x, a), a, y) → C(y, a, f(c(z, y, x)))
The TRS R consists of the following rules:
b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))
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:
F(f(c(a, y, z))) → B(y, b(z, z))
C(c(z, x, a), a, y) → F(f(c(y, a, f(c(z, y, x)))))
C(c(z, x, a), a, y) → F(c(z, y, x))
F(f(c(a, y, z))) → B(z, z)
C(c(z, x, a), a, y) → F(c(y, a, f(c(z, y, x))))
C(c(z, x, a), a, y) → C(z, y, x)
C(c(z, x, a), a, y) → C(y, a, f(c(z, y, x)))
The TRS R consists of the following rules:
b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))
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:
C(c(z, x, a), a, y) → F(f(c(y, a, f(c(z, y, x)))))
F(f(c(a, y, z))) → B(y, b(z, z))
C(c(z, x, a), a, y) → F(c(z, y, x))
C(c(z, x, a), a, y) → F(c(y, a, f(c(z, y, x))))
F(f(c(a, y, z))) → B(z, z)
C(c(z, x, a), a, y) → C(y, a, f(c(z, y, x)))
C(c(z, x, a), a, y) → C(z, y, x)
The TRS R consists of the following rules:
b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(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 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
C(c(z, x, a), a, y) → C(y, a, f(c(z, y, x)))
C(c(z, x, a), a, y) → C(z, y, x)
The TRS R consists of the following rules:
b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.