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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

B(x, b(z, y)) → F(z)
B(x, b(z, y)) → B(f(f(z)), c(x, z, y))
B(x, b(z, y)) → F(b(f(f(z)), c(x, z, y)))
F(c(a, z, x)) → B(a, z)
B(x, b(z, y)) → F(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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

B(x, b(z, y)) → F(z)
B(x, b(z, y)) → F(b(f(f(z)), c(x, z, y)))
F(c(a, z, x)) → B(a, z)
B(x, b(z, y)) → F(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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


B(x, b(z, y)) → F(z)
B(x, b(z, y)) → F(f(z))
The remaining pairs can at least be oriented weakly.

B(x, b(z, y)) → F(b(f(f(z)), c(x, z, y)))
F(c(a, z, x)) → B(a, z)
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1)) = x_1   
POL(a) = 0   
POL(b(x1, x2)) = 1 + (4)x_1 + (2)x_2   
POL(c(x1, x2, x3)) = 1 + (2)x_2 + (2)x_3   
POL(B(x1, x2)) = 1 + (2)x_2   
POL(F(x1)) = x_1   
The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP

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

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