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(a, z, x)) → b(a, z)
b(x, b(z, y)) → f(b(f(f(z)), c(x, z, y)))
b(y, z) → z

Q is empty.


QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f(c(a, z, x)) → b(a, z)
b(x, b(z, y)) → f(b(f(f(z)), c(x, z, y)))
b(y, 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(a, z, x)) → B(a, z)
B(x, b(z, y)) → F(b(f(f(z)), c(x, z, y)))
B(x, b(z, y)) → F(f(z))
B(x, b(z, y)) → F(z)
B(x, b(z, y)) → B(f(f(z)), c(x, z, y))

The TRS R consists of the following rules:

f(c(a, z, x)) → b(a, z)
b(x, b(z, y)) → f(b(f(f(z)), c(x, z, y)))
b(y, 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(a, z, x)) → B(a, z)
B(x, b(z, y)) → F(b(f(f(z)), c(x, z, y)))
B(x, b(z, y)) → F(f(z))
B(x, b(z, y)) → F(z)
B(x, b(z, y)) → B(f(f(z)), c(x, z, y))

The TRS R consists of the following rules:

f(c(a, z, x)) → b(a, z)
b(x, b(z, y)) → f(b(f(f(z)), c(x, z, y)))
b(y, 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(a, z, x)) → B(a, z)
B(x, b(z, y)) → F(b(f(f(z)), c(x, z, y)))
B(x, b(z, y)) → F(f(z))
B(x, b(z, y)) → F(z)
B(x, b(z, y)) → B(f(f(z)), c(x, z, y))

The TRS R consists of the following rules:

f(c(a, z, x)) → b(a, z)
b(x, b(z, y)) → f(b(f(f(z)), c(x, z, y)))
b(y, 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 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
QDP

Q DP problem:
The TRS P consists of the following rules:

F(c(a, z, x)) → B(a, z)
B(x, b(z, y)) → F(b(f(f(z)), c(x, z, y)))
B(x, b(z, y)) → F(f(z))
B(x, b(z, y)) → F(z)

The TRS R consists of the following rules:

f(c(a, z, x)) → b(a, z)
b(x, b(z, y)) → f(b(f(f(z)), c(x, z, y)))
b(y, z) → z

Q is empty.
We have to consider all minimal (P,Q,R)-chains.