The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(1)).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
s(log(0)) → s(0)
log(s(x)) → s(log(half(s(x))))

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0
half(s(0)) → 0
half(s(s(z0))) → s(half(z0))
s(log(0)) → s(0)
log(s(z0)) → s(log(half(s(z0))))

Tuples:

HALF(s(s(z0))) → c2(S(half(z0)), HALF(z0))
S(log(0)) → c3(S(0))
LOG(s(z0)) → c4(S(log(half(s(z0)))), LOG(half(s(z0))), HALF(s(z0)), S(z0))

S tuples:

HALF(s(s(z0))) → c2(S(half(z0)), HALF(z0))
S(log(0)) → c3(S(0))
LOG(s(z0)) → c4(S(log(half(s(z0)))), LOG(half(s(z0))), HALF(s(z0)), S(z0))

K tuples:none
Defined Rule Symbols:

half, s, log

Defined Pair Symbols:

HALF, S, LOG

Compound Symbols:

c2, c3, c4

(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

HALF(s(s(z0))) → c2(S(half(z0)), HALF(z0))
S(log(0)) → c3(S(0))
LOG(s(z0)) → c4(S(log(half(s(z0)))), LOG(half(s(z0))), HALF(s(z0)), S(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

half(0) → 0
half(s(0)) → 0
half(s(s(z0))) → s(half(z0))
s(log(0)) → s(0)
log(s(z0)) → s(log(half(s(z0))))

Tuples:none
S tuples:none
K tuples:none
Defined Rule Symbols:

half, s, log

Defined Pair Symbols:none

Compound Symbols:none

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(6) BOUNDS(O(1), O(1))