The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^3)).

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
↳2 CdtProblem
↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
↳4 CdtProblem
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^3))))
↳6 CdtProblem
↳7 SIsEmptyProof (⇔)
↳8 BOUNDS(O(1), O(1))

(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^1))) transformation)

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

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) = [3]   
POL(HALF(x1)) = 0   
POL(LOG(x1)) = x1   
POL(c1(x1)) = x1   
POL(c3(x1, x2)) = x1 + x2   
POL(half(x1)) = [1] + x1   
POL(s(x1)) = [2] + 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:

HALF(s(s(z0))) → c1(HALF(z0))
K tuples:

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) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^3))) 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))
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)) = x13   
POL(c1(x1)) = x1   
POL(c3(x1, x2)) = x1 + x2   
POL(half(x1)) = x1   
POL(s(x1)) = [1] + x1   

(6) 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:

LOG(s(s(z0))) → c3(LOG(s(half(z0))), HALF(z0))
HALF(s(s(z0))) → c1(HALF(z0))
Defined Rule Symbols:

half, log

Defined Pair Symbols:

HALF, LOG

Compound Symbols:

c1, c3

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))