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:

c3(c3(z, y, a), a, a) -> b2(z, y)
f1(c3(x, y, z)) -> c3(z, f1(b2(y, z)), a)
b2(z, b2(c3(a, y, a), f1(f1(x)))) -> c3(c3(y, a, z), z, x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

c3(c3(z, y, a), a, a) -> b2(z, y)
f1(c3(x, y, z)) -> c3(z, f1(b2(y, z)), a)
b2(z, b2(c3(a, y, a), f1(f1(x)))) -> c3(c3(y, a, z), z, x)

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:

B2(z, b2(c3(a, y, a), f1(f1(x)))) -> C3(y, a, z)
F1(c3(x, y, z)) -> B2(y, z)
F1(c3(x, y, z)) -> C3(z, f1(b2(y, z)), a)
C3(c3(z, y, a), a, a) -> B2(z, y)
B2(z, b2(c3(a, y, a), f1(f1(x)))) -> C3(c3(y, a, z), z, x)
F1(c3(x, y, z)) -> F1(b2(y, z))

The TRS R consists of the following rules:

c3(c3(z, y, a), a, a) -> b2(z, y)
f1(c3(x, y, z)) -> c3(z, f1(b2(y, z)), a)
b2(z, b2(c3(a, y, a), f1(f1(x)))) -> c3(c3(y, a, z), z, x)

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:

B2(z, b2(c3(a, y, a), f1(f1(x)))) -> C3(y, a, z)
F1(c3(x, y, z)) -> B2(y, z)
F1(c3(x, y, z)) -> C3(z, f1(b2(y, z)), a)
C3(c3(z, y, a), a, a) -> B2(z, y)
B2(z, b2(c3(a, y, a), f1(f1(x)))) -> C3(c3(y, a, z), z, x)
F1(c3(x, y, z)) -> F1(b2(y, z))

The TRS R consists of the following rules:

c3(c3(z, y, a), a, a) -> b2(z, y)
f1(c3(x, y, z)) -> c3(z, f1(b2(y, z)), a)
b2(z, b2(c3(a, y, a), f1(f1(x)))) -> c3(c3(y, a, z), z, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 2 less nodes.

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

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

B2(z, b2(c3(a, y, a), f1(f1(x)))) -> C3(y, a, z)
C3(c3(z, y, a), a, a) -> B2(z, y)
B2(z, b2(c3(a, y, a), f1(f1(x)))) -> C3(c3(y, a, z), z, x)

The TRS R consists of the following rules:

c3(c3(z, y, a), a, a) -> b2(z, y)
f1(c3(x, y, z)) -> c3(z, f1(b2(y, z)), a)
b2(z, b2(c3(a, y, a), f1(f1(x)))) -> c3(c3(y, a, z), z, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


B2(z, b2(c3(a, y, a), f1(f1(x)))) -> C3(y, a, z)
C3(c3(z, y, a), a, a) -> B2(z, y)
B2(z, b2(c3(a, y, a), f1(f1(x)))) -> C3(c3(y, a, z), z, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(B2(x1, x2)) = x2   
POL(C3(x1, x2, x3)) = x1   
POL(a) = 1   
POL(b2(x1, x2)) = x1 + x2   
POL(c3(x1, x2, x3)) = 1 + x1 + x2   
POL(f1(x1)) = 1   

The following usable rules [14] were oriented:

b2(z, b2(c3(a, y, a), f1(f1(x)))) -> c3(c3(y, a, z), z, x)
c3(c3(z, y, a), a, a) -> b2(z, y)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

c3(c3(z, y, a), a, a) -> b2(z, y)
f1(c3(x, y, z)) -> c3(z, f1(b2(y, z)), a)
b2(z, b2(c3(a, y, a), f1(f1(x)))) -> c3(c3(y, a, z), z, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP

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

F1(c3(x, y, z)) -> F1(b2(y, z))

The TRS R consists of the following rules:

c3(c3(z, y, a), a, a) -> b2(z, y)
f1(c3(x, y, z)) -> c3(z, f1(b2(y, z)), a)
b2(z, b2(c3(a, y, a), f1(f1(x)))) -> c3(c3(y, a, z), z, x)

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