(0) Obligation:
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.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(c(c(a(x, y)))) → C(c(c(c(y))))
C(c(c(a(x, y)))) → C(c(c(y)))
C(c(c(a(x, y)))) → C(c(y))
C(c(c(a(x, y)))) → C(y)
C(c(b(c(y), 0))) → C(c(a(y, 0)))
C(c(b(c(y), 0))) → C(a(y, 0))
C(c(a(a(y, 0), x))) → 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.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(c(c(a(x, y)))) → C(c(c(y)))
C(c(c(a(x, y)))) → C(c(c(c(y))))
C(c(c(a(x, y)))) → C(c(y))
C(c(c(a(x, y)))) → C(y)
C(c(b(c(y), 0))) → C(c(a(y, 0)))
C(c(a(a(y, 0), x))) → 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.