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