(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(0, y), x) → y
c(c(c(y))) → c(c(a(a(c(b(0, y)), 0), 0)))
a(y, 0) → b(y, 0)
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(y))) → C(c(a(a(c(b(0, y)), 0), 0)))
C(c(c(y))) → C(a(a(c(b(0, y)), 0), 0))
C(c(c(y))) → A(a(c(b(0, y)), 0), 0)
C(c(c(y))) → A(c(b(0, y)), 0)
C(c(c(y))) → C(b(0, y))
C(c(c(y))) → B(0, y)
A(y, 0) → B(y, 0)
The TRS R consists of the following rules:
b(b(0, y), x) → y
c(c(c(y))) → c(c(a(a(c(b(0, y)), 0), 0)))
a(y, 0) → b(y, 0)
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 5 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(c(c(y))) → C(a(a(c(b(0, y)), 0), 0))
C(c(c(y))) → C(c(a(a(c(b(0, y)), 0), 0)))
The TRS R consists of the following rules:
b(b(0, y), x) → y
c(c(c(y))) → c(c(a(a(c(b(0, y)), 0), 0)))
a(y, 0) → b(y, 0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.