The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

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

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

Tuples:

HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))

S tuples:

HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))

K tuples:none
Defined Rule Symbols:

half, log

Defined Pair Symbols:

HALF, LOG

Compound Symbols:

c1, c3

(3) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))

We considered the (Usable) Rules:

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

And the Tuples:

HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [1]   
POL(HALF(x₁)) = x₁   
POL(LOG(x₁)) = x₁²   
POL(c1(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(half(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))

S tuples:none
K tuples:

HALF(s(s(z0))) → c1(HALF(z0))
LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))

Defined Rule Symbols:

half, log

Defined Pair Symbols:

HALF, LOG

Compound Symbols:

c1, c3

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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