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)) → +(sum(x), s(x))
sum1(0) → 0
sum1(s(x)) → s(+(sum1(x), +(x, 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)) → +(sum(z0), s(z0))
sum1(0) → 0
sum1(s(z0)) → s(+(sum1(z0), +(z0, z0)))

Tuples:

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))

S tuples:

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))

K tuples:none
Defined Rule Symbols:

sum, sum1

Defined Pair Symbols:

SUM, SUM1

Compound Symbols:

c1, 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.

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))

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

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))

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

POL(SUM(x₁)) = [4]x₁   
POL(SUM1(x₁)) = [4]x₁   
POL(c1(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(s(x₁)) = [4] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
sum1(0) → 0
sum1(s(z0)) → s(+(sum1(z0), +(z0, z0)))

Tuples:

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))

S tuples:none
K tuples:

SUM(s(z0)) → c1(SUM(z0))
SUM1(s(z0)) → c3(SUM1(z0))

Defined Rule Symbols:

sum, sum1

Defined Pair Symbols:

SUM, SUM1

Compound Symbols:

c1, c3

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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