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(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → z
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → 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(z, x, a) → F(b(b(f(z), z), x))
C(z, x, a) → B(f(z), z)
B(y, b(z, a)) → B(c(f(a), y, z), z)
B(y, b(z, a)) → C(f(a), y, z)
C(z, x, a) → B(b(f(z), z), x)
B(y, b(z, a)) → F(a)
B(y, b(z, a)) → F(b(c(f(a), y, z), z))
C(z, x, a) → F(z)
The TRS R consists of the following rules:
c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → 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(z, x, a) → F(b(b(f(z), z), x))
C(z, x, a) → B(f(z), z)
B(y, b(z, a)) → B(c(f(a), y, z), z)
B(y, b(z, a)) → C(f(a), y, z)
C(z, x, a) → B(b(f(z), z), x)
B(y, b(z, a)) → F(a)
B(y, b(z, a)) → F(b(c(f(a), y, z), z))
C(z, x, a) → F(z)
The TRS R consists of the following rules:
c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → 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(z, x, a) → F(b(b(f(z), z), x))
C(z, x, a) → B(f(z), z)
B(y, b(z, a)) → B(c(f(a), y, z), z)
C(z, x, a) → B(b(f(z), z), x)
B(y, b(z, a)) → C(f(a), y, z)
B(y, b(z, a)) → F(a)
B(y, b(z, a)) → F(b(c(f(a), y, z), z))
C(z, x, a) → F(z)
The TRS R consists of the following rules:
c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → 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 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
C(z, x, a) → B(f(z), z)
B(y, b(z, a)) → B(c(f(a), y, z), z)
B(y, b(z, a)) → C(f(a), y, z)
C(z, x, a) → B(b(f(z), z), x)
The TRS R consists of the following rules:
c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → z
Q is empty.
We have to consider all minimal (P,Q,R)-chains.