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:

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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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(a2(x, y)))) -> C1(c1(c1(c1(y))))
C1(c1(c1(a2(x, y)))) -> C1(y)
C1(c1(c1(a2(x, y)))) -> C1(c1(c1(y)))
C1(c1(a2(a2(y, 0), x))) -> C1(y)
C1(c1(c1(a2(x, y)))) -> C1(c1(y))
C1(c1(b2(c1(y), 0))) -> C1(c1(a2(y, 0)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(C1(x1)) = 2·x12   
POL(a2(x1, x2)) = 1 + x1 + 2·x2   
POL(b2(x1, x2)) = 3 + x1   
POL(c1(x1)) = 2·x1   

The following usable rules [14] were oriented:

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



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

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

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

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.