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:

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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

c(c(c(y))) → c(c(a(y, 0)))
c(a(a(0, x), y)) → a(c(c(c(0))), y)
c(y) → 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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
QDP

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

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

The TRS R consists of the following rules:

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

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