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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

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

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