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:

*(x, *(y, z)) → *(*(x, y), z)
*(x, x) → x

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

*

Defined Pair Symbols:

*'

Compound Symbols:

c

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

We considered the (Usable) Rules:

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

And the Tuples:

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

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

POL(*(x₁, x₂)) = [2] + x₁ + [3]x₂   
POL(*'(x₁, x₂)) = x₂   
POL(c(x₁, x₂)) = x₁ + x₂   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:none
K tuples:

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

Defined Rule Symbols:

*

Defined Pair Symbols:

*'

Compound Symbols:

c

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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