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:

sum(0) → 0
sum(s(x)) → +(sqr(s(x)), sum(x))
sqr(x) → *(x, x)
sum(s(x)) → +(*(s(x), s(x)), sum(x))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sqr(s(z0)), sum(z0))
sum(s(z0)) → +(*(s(z0), s(z0)), sum(z0))
sqr(z0) → *(z0, z0)

Tuples:

SUM(s(z0)) → c1(SQR(s(z0)), SUM(z0))
SUM(s(z0)) → c2(SUM(z0))

S tuples:

SUM(s(z0)) → c1(SQR(s(z0)), SUM(z0))
SUM(s(z0)) → c2(SUM(z0))

K tuples:none
Defined Rule Symbols:

sum, sqr

Defined Pair Symbols:

SUM

Compound Symbols:

c1, c2

(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.

SUM(s(z0)) → c1(SQR(s(z0)), SUM(z0))
SUM(s(z0)) → c2(SUM(z0))

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

SUM(s(z0)) → c1(SQR(s(z0)), SUM(z0))
SUM(s(z0)) → c2(SUM(z0))

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

POL(SQR(x₁)) = 0   
POL(SUM(x₁)) = [4]x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁)) = x₁   
POL(s(x₁)) = [3] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sqr(s(z0)), sum(z0))
sum(s(z0)) → +(*(s(z0), s(z0)), sum(z0))
sqr(z0) → *(z0, z0)

Tuples:

SUM(s(z0)) → c1(SQR(s(z0)), SUM(z0))
SUM(s(z0)) → c2(SUM(z0))

S tuples:none
K tuples:

SUM(s(z0)) → c1(SQR(s(z0)), SUM(z0))
SUM(s(z0)) → c2(SUM(z0))

Defined Rule Symbols:

sum, sqr

Defined Pair Symbols:

SUM

Compound Symbols:

c1, c2

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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