(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
bits(0)
bits(s(x0))
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x))) → HALF(x)
BITS(s(x)) → BITS(half(s(x)))
BITS(s(x)) → HALF(s(x))
The TRS R consists of the following rules:
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
bits(0)
bits(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x))) → HALF(x)
The TRS R consists of the following rules:
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
bits(0)
bits(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
BITS(s(x)) → BITS(half(s(x)))
The TRS R consists of the following rules:
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))
The set Q consists of the following terms:
half(0)
half(s(0))
half(s(s(x0)))
bits(0)
bits(s(x0))
We have to consider all minimal (P,Q,R)-chains.