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, *(minus(y), y)) → *(minus(*(y, y)), x)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

*(z0, *(minus(z1), z1)) → *(minus(*(z1, z1)), z0)

Tuples:

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

S tuples:

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

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

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

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

POL(*(x₁, x₂)) = [3] + [5]x₁ + [4]x₂   
POL(*'(x₁, x₂)) = [3] + [4]x₁ + [3]x₂   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁)) = [5]   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

*(z0, *(minus(z1), z1)) → *(minus(*(z1, z1)), z0)

Tuples:

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

S tuples:none
K tuples:

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