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(c(c(a(x, y)))) → b(c(c(c(c(y)))), x)
c(c(b(c(y), 0))) → a(0, c(c(a(y, 0))))
c(c(a(a(y, 0), x))) → c(y)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

c(c(c(a(x, y)))) → b(c(c(c(c(y)))), x)
c(c(b(c(y), 0))) → a(0, c(c(a(y, 0))))
c(c(a(a(y, 0), x))) → c(y)

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(c(c(a(x, y)))) → C(y)
C(c(a(a(y, 0), x))) → C(y)
C(c(c(a(x, y)))) → C(c(y))
C(c(b(c(y), 0))) → C(a(y, 0))
C(c(b(c(y), 0))) → C(c(a(y, 0)))
C(c(c(a(x, y)))) → C(c(c(y)))
C(c(c(a(x, y)))) → C(c(c(c(y))))

The TRS R consists of the following rules:

c(c(c(a(x, y)))) → b(c(c(c(c(y)))), x)
c(c(b(c(y), 0))) → a(0, c(c(a(y, 0))))
c(c(a(a(y, 0), x))) → c(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:

C(c(c(a(x, y)))) → C(y)
C(c(a(a(y, 0), x))) → C(y)
C(c(c(a(x, y)))) → C(c(y))
C(c(b(c(y), 0))) → C(a(y, 0))
C(c(b(c(y), 0))) → C(c(a(y, 0)))
C(c(c(a(x, y)))) → C(c(c(y)))
C(c(c(a(x, y)))) → C(c(c(c(y))))

The TRS R consists of the following rules:

c(c(c(a(x, y)))) → b(c(c(c(c(y)))), x)
c(c(b(c(y), 0))) → a(0, c(c(a(y, 0))))
c(c(a(a(y, 0), x))) → c(y)

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 1 less node.

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

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

C(c(c(a(x, y)))) → C(y)
C(c(a(a(y, 0), x))) → C(y)
C(c(c(a(x, y)))) → C(c(y))
C(c(b(c(y), 0))) → C(c(a(y, 0)))
C(c(c(a(x, y)))) → C(c(c(y)))
C(c(c(a(x, y)))) → C(c(c(c(y))))

The TRS R consists of the following rules:

c(c(c(a(x, y)))) → b(c(c(c(c(y)))), x)
c(c(b(c(y), 0))) → a(0, c(c(a(y, 0))))
c(c(a(a(y, 0), x))) → c(y)

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(c(c(a(x, y)))) → C(y)
C(c(a(a(y, 0), x))) → C(y)
C(c(c(a(x, y)))) → C(c(y))
C(c(b(c(y), 0))) → C(c(a(y, 0)))
C(c(c(a(x, y)))) → C(c(c(y)))
C(c(c(a(x, y)))) → C(c(c(c(y))))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(C(x1)) = (4)x_1   
POL(c(x1)) = (4)x_1   
POL(b(x1, x2)) = 4 + x_1   
POL(a(x1, x2)) = 1/4 + x_1 + (4)x_2   
POL(0) = 0   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:

c(c(c(a(x, y)))) → b(c(c(c(c(y)))), x)
c(c(b(c(y), 0))) → a(0, c(c(a(y, 0))))
c(c(a(a(y, 0), x))) → c(y)



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

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

c(c(c(a(x, y)))) → b(c(c(c(c(y)))), x)
c(c(b(c(y), 0))) → a(0, c(c(a(y, 0))))
c(c(a(a(y, 0), x))) → c(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.