Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

C(z, x, a) → F(z)
C(z, x, a) → B(f(z), z)
B(y, b(z, a)) → C(f(a), y, z)
B(y, b(z, a)) → F(a)
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)) → F(b(c(f(a), y, z), z))
C(z, x, a) → F(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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

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

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

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


The following pairs can be oriented strictly and are deleted.


C(z, x, a) → B(f(z), z)
The remaining pairs can at least be oriented weakly.

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

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

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



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

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

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


The following pairs can be oriented strictly and are deleted.


B(y, b(z, a)) → C(f(a), y, z)
B(y, b(z, a)) → B(c(f(a), y, z), z)
The remaining pairs can at least be oriented weakly.

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

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

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



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

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

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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.