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:

b2(b2(0, y), x) -> y
c1(c1(c1(y))) -> c1(c1(a2(a2(c1(b2(0, y)), 0), 0)))
a2(y, 0) -> b2(y, 0)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

b2(b2(0, y), x) -> y
c1(c1(c1(y))) -> c1(c1(a2(a2(c1(b2(0, y)), 0), 0)))
a2(y, 0) -> b2(y, 0)

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(y))) -> A2(c1(b2(0, y)), 0)
C1(c1(c1(y))) -> C1(b2(0, y))
A2(y, 0) -> B2(y, 0)
C1(c1(c1(y))) -> C1(a2(a2(c1(b2(0, y)), 0), 0))
C1(c1(c1(y))) -> C1(c1(a2(a2(c1(b2(0, y)), 0), 0)))
C1(c1(c1(y))) -> B2(0, y)
C1(c1(c1(y))) -> A2(a2(c1(b2(0, y)), 0), 0)

The TRS R consists of the following rules:

b2(b2(0, y), x) -> y
c1(c1(c1(y))) -> c1(c1(a2(a2(c1(b2(0, y)), 0), 0)))
a2(y, 0) -> b2(y, 0)

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:

C1(c1(c1(y))) -> A2(c1(b2(0, y)), 0)
C1(c1(c1(y))) -> C1(b2(0, y))
A2(y, 0) -> B2(y, 0)
C1(c1(c1(y))) -> C1(a2(a2(c1(b2(0, y)), 0), 0))
C1(c1(c1(y))) -> C1(c1(a2(a2(c1(b2(0, y)), 0), 0)))
C1(c1(c1(y))) -> B2(0, y)
C1(c1(c1(y))) -> A2(a2(c1(b2(0, y)), 0), 0)

The TRS R consists of the following rules:

b2(b2(0, y), x) -> y
c1(c1(c1(y))) -> c1(c1(a2(a2(c1(b2(0, y)), 0), 0)))
a2(y, 0) -> b2(y, 0)

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 5 less nodes.

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

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

C1(c1(c1(y))) -> C1(a2(a2(c1(b2(0, y)), 0), 0))
C1(c1(c1(y))) -> C1(c1(a2(a2(c1(b2(0, y)), 0), 0)))

The TRS R consists of the following rules:

b2(b2(0, y), x) -> y
c1(c1(c1(y))) -> c1(c1(a2(a2(c1(b2(0, y)), 0), 0)))
a2(y, 0) -> b2(y, 0)

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(y))) -> C1(a2(a2(c1(b2(0, y)), 0), 0))
The remaining pairs can at least be oriented weakly.

C1(c1(c1(y))) -> C1(c1(a2(a2(c1(b2(0, y)), 0), 0)))
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(C1(x1)) = x1   
POL(a2(x1, x2)) = x1   
POL(b2(x1, x2)) = x1 + x2   
POL(c1(x1)) = 2 + x1   

The following usable rules [14] were oriented:

a2(y, 0) -> b2(y, 0)
b2(b2(0, y), x) -> y
c1(c1(c1(y))) -> c1(c1(a2(a2(c1(b2(0, y)), 0), 0)))



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

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

C1(c1(c1(y))) -> C1(c1(a2(a2(c1(b2(0, y)), 0), 0)))

The TRS R consists of the following rules:

b2(b2(0, y), x) -> y
c1(c1(c1(y))) -> c1(c1(a2(a2(c1(b2(0, y)), 0), 0)))
a2(y, 0) -> b2(y, 0)

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