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:

c1(c1(c1(b1(x)))) -> a2(1, b1(c1(x)))
b1(c1(b1(c1(x)))) -> a2(0, a2(1, x))
a2(0, x) -> c1(c1(x))
a2(1, x) -> c1(b1(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

c1(c1(c1(b1(x)))) -> a2(1, b1(c1(x)))
b1(c1(b1(c1(x)))) -> a2(0, a2(1, x))
a2(0, x) -> c1(c1(x))
a2(1, x) -> c1(b1(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:

C1(c1(c1(b1(x)))) -> C1(x)
A2(1, x) -> C1(b1(x))
B1(c1(b1(c1(x)))) -> A2(1, x)
A2(1, x) -> B1(x)
C1(c1(c1(b1(x)))) -> A2(1, b1(c1(x)))
B1(c1(b1(c1(x)))) -> A2(0, a2(1, x))
A2(0, x) -> C1(x)
A2(0, x) -> C1(c1(x))
C1(c1(c1(b1(x)))) -> B1(c1(x))

The TRS R consists of the following rules:

c1(c1(c1(b1(x)))) -> a2(1, b1(c1(x)))
b1(c1(b1(c1(x)))) -> a2(0, a2(1, x))
a2(0, x) -> c1(c1(x))
a2(1, x) -> c1(b1(x))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

C1(c1(c1(b1(x)))) -> C1(x)
A2(1, x) -> C1(b1(x))
B1(c1(b1(c1(x)))) -> A2(1, x)
A2(1, x) -> B1(x)
C1(c1(c1(b1(x)))) -> A2(1, b1(c1(x)))
B1(c1(b1(c1(x)))) -> A2(0, a2(1, x))
A2(0, x) -> C1(x)
A2(0, x) -> C1(c1(x))
C1(c1(c1(b1(x)))) -> B1(c1(x))

The TRS R consists of the following rules:

c1(c1(c1(b1(x)))) -> a2(1, b1(c1(x)))
b1(c1(b1(c1(x)))) -> a2(0, a2(1, x))
a2(0, x) -> c1(c1(x))
a2(1, x) -> c1(b1(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.


C1(c1(c1(b1(x)))) -> C1(x)
B1(c1(b1(c1(x)))) -> A2(1, x)
A2(1, x) -> B1(x)
A2(0, x) -> C1(x)
C1(c1(c1(b1(x)))) -> B1(c1(x))
The remaining pairs can at least be oriented weakly.

A2(1, x) -> C1(b1(x))
C1(c1(c1(b1(x)))) -> A2(1, b1(c1(x)))
B1(c1(b1(c1(x)))) -> A2(0, a2(1, x))
A2(0, x) -> C1(c1(x))
Used ordering: Polynomial interpretation [21]:

POL(0) = 1   
POL(1) = 1   
POL(A2(x1, x2)) = x1 + x2   
POL(B1(x1)) = x1   
POL(C1(x1)) = x1   
POL(a2(x1, x2)) = 2·x1 + x2   
POL(b1(x1)) = 1 + x1   
POL(c1(x1)) = 1 + x1   

The following usable rules [14] were oriented:

a2(0, x) -> c1(c1(x))
b1(c1(b1(c1(x)))) -> a2(0, a2(1, x))
a2(1, x) -> c1(b1(x))
c1(c1(c1(b1(x)))) -> a2(1, b1(c1(x)))



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

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

A2(1, x) -> C1(b1(x))
C1(c1(c1(b1(x)))) -> A2(1, b1(c1(x)))
B1(c1(b1(c1(x)))) -> A2(0, a2(1, x))
A2(0, x) -> C1(c1(x))

The TRS R consists of the following rules:

c1(c1(c1(b1(x)))) -> a2(1, b1(c1(x)))
b1(c1(b1(c1(x)))) -> a2(0, a2(1, x))
a2(0, x) -> c1(c1(x))
a2(1, x) -> c1(b1(x))

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 2 less nodes.

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

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

A2(1, x) -> C1(b1(x))
C1(c1(c1(b1(x)))) -> A2(1, b1(c1(x)))

The TRS R consists of the following rules:

c1(c1(c1(b1(x)))) -> a2(1, b1(c1(x)))
b1(c1(b1(c1(x)))) -> a2(0, a2(1, x))
a2(0, x) -> c1(c1(x))
a2(1, x) -> c1(b1(x))

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