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

(0) Obligation:

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

*(i(x), x) → 1
*(1, y) → y
*(x, 0) → 0
*(*(x, y), z) → *(x, *(y, z))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

*(i(z0), z0) → 1
*(1, z0) → z0
*(z0, 0) → 0
*(*(z0, z1), z2) → *(z0, *(z1, z2))

Tuples:

*'(*(z0, z1), z2) → c3(*'(z0, *(z1, z2)), *'(z1, z2))

S tuples:

*'(*(z0, z1), z2) → c3(*'(z0, *(z1, z2)), *'(z1, z2))

K tuples:none
Defined Rule Symbols:

*

Defined Pair Symbols:

*'

Compound Symbols:

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.

*'(*(z0, z1), z2) → c3(*'(z0, *(z1, z2)), *'(z1, z2))

We considered the (Usable) Rules:

*(i(z0), z0) → 1
*(1, z0) → z0
*(z0, 0) → 0
*(*(z0, z1), z2) → *(z0, *(z1, z2))

And the Tuples:

*'(*(z0, z1), z2) → c3(*'(z0, *(z1, z2)), *'(z1, z2))

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

POL(*(x₁, x₂)) = [4] + [4]x₁ + [4]x₂   
POL(*'(x₁, x₂)) = [4] + [4]x₁   
POL(0) = [4]   
POL(1) = [3]   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(i(x₁)) = [2]   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

*(i(z0), z0) → 1
*(1, z0) → z0
*(z0, 0) → 0
*(*(z0, z1), z2) → *(z0, *(z1, z2))

Tuples:

*'(*(z0, z1), z2) → c3(*'(z0, *(z1, z2)), *'(z1, z2))

S tuples:none
K tuples:

*'(*(z0, z1), z2) → c3(*'(z0, *(z1, z2)), *'(z1, z2))

Defined Rule Symbols:

*

Defined Pair Symbols:

*'

Compound Symbols:

c3

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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