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(f(b(x, z)), y) → f(f(f(b(z, b(y, z)))))
c(f(f(c(x, a, z))), a, y) → b(y, f(b(a, z)))
b(b(c(b(a, a), a, z), f(a)), y) → z
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(f(b(x, z)), y) → f(f(f(b(z, b(y, z)))))
c(f(f(c(x, a, z))), a, y) → b(y, f(b(a, z)))
b(b(c(b(a, a), a, z), f(a)), y) → 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:
C(f(f(c(x, a, z))), a, y) → B(a, z)
C(f(f(c(x, a, z))), a, y) → B(y, f(b(a, z)))
B(f(b(x, z)), y) → B(z, b(y, z))
B(f(b(x, z)), y) → B(y, z)
The TRS R consists of the following rules:
b(f(b(x, z)), y) → f(f(f(b(z, b(y, z)))))
c(f(f(c(x, a, z))), a, y) → b(y, f(b(a, z)))
b(b(c(b(a, a), a, z), f(a)), y) → 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:
C(f(f(c(x, a, z))), a, y) → B(a, z)
C(f(f(c(x, a, z))), a, y) → B(y, f(b(a, z)))
B(f(b(x, z)), y) → B(z, b(y, z))
B(f(b(x, z)), y) → B(y, z)
The TRS R consists of the following rules:
b(f(b(x, z)), y) → f(f(f(b(z, b(y, z)))))
c(f(f(c(x, a, z))), a, y) → b(y, f(b(a, z)))
b(b(c(b(a, a), a, z), f(a)), y) → 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(f(f(c(x, a, z))), a, y) → B(a, z)
C(f(f(c(x, a, z))), a, y) → B(y, f(b(a, z)))
B(f(b(x, z)), y) → B(z, b(y, z))
B(f(b(x, z)), y) → B(y, z)
The TRS R consists of the following rules:
b(f(b(x, z)), y) → f(f(f(b(z, b(y, z)))))
c(f(f(c(x, a, z))), a, y) → b(y, f(b(a, z)))
b(b(c(b(a, a), a, z), f(a)), y) → 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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(f(b(x, z)), y) → B(z, b(y, z))
B(f(b(x, z)), y) → B(y, z)
The TRS R consists of the following rules:
b(f(b(x, z)), y) → f(f(f(b(z, b(y, z)))))
c(f(f(c(x, a, z))), a, y) → b(y, f(b(a, z)))
b(b(c(b(a, a), a, z), f(a)), y) → z
Q is empty.
We have to consider all minimal (P,Q,R)-chains.