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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

a(a(y, 0), 0) → y
c(c(y)) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), 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 2 less nodes.

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

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

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

The TRS R consists of the following rules:

a(a(y, 0), 0) → y
c(c(y)) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), 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(a(c(c(y)), x)) → C(c(a(x, 0)))
C(a(c(c(y)), x)) → C(a(x, 0))
The remaining pairs can at least be oriented weakly.

C(a(c(c(y)), x)) → C(c(c(a(x, 0))))
Used ordering: Polynomial interpretation [25,35]:

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

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



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

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

C(a(c(c(y)), x)) → C(c(c(a(x, 0))))

The TRS R consists of the following rules:

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

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