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), *(x, z))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:

*

Defined Pair Symbols:

*'

Compound Symbols:

c

(3) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

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

S tuples:

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

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

*'

Compound Symbols:

c

(5) CdtRuleRemovalProof (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), *'(z0, z2))

We considered the (Usable) Rules:none
And the Tuples:

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

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

POL(*'(x₁, x₂)) = [3] + [4]x₂   
POL(+(x₁, x₂)) = [4] + x₁ + x₂   
POL(c(x₁, x₂)) = x₁ + x₂   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

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

S tuples:none
K tuples:

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

Defined Rule Symbols:none

Defined Pair Symbols:

*'

Compound Symbols:

c

(7) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

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