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