The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

sum(0) → 0
sum(s(x)) → +(sum(x), s(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

S tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

K tuples:none
Defined Rule Symbols:

sum, +

Defined Pair Symbols:

SUM, +'

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), s(z0)), SUM(z0))

We considered the (Usable) Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
+(z0, s(z1)) → s(+(z0, z1))
+(z0, 0) → z0

And the Tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

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

POL(+(x₁, x₂)) = [4] + [2]x₁   
POL(+'(x₁, x₂)) = [2]   
POL(0) = [5]   
POL(SUM(x₁)) = [4]x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(s(x₁)) = [5] + x₁   
POL(sum(x₁)) = [2]   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

S tuples:

+'(z0, s(z1)) → c3(+'(z0, z1))

K tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))

Defined Rule Symbols:

sum, +

Defined Pair Symbols:

SUM, +'

Compound Symbols:

c1, c3

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

+'(z0, s(z1)) → c3(+'(z0, z1))

We considered the (Usable) Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
+(z0, s(z1)) → s(+(z0, z1))
+(z0, 0) → z0

And the Tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

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

POL(+(x₁, x₂)) = 0   
POL(+'(x₁, x₂)) = [1] + [2]x₂   
POL(0) = 0   
POL(SUM(x₁)) = [3]x₁ + x₁²   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   
POL(sum(x₁)) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

sum(0) → 0
sum(s(z0)) → +(sum(z0), s(z0))
+(z0, 0) → z0
+(z0, s(z1)) → s(+(z0, z1))

Tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

S tuples:none
K tuples:

SUM(s(z0)) → c1(+'(sum(z0), s(z0)), SUM(z0))
+'(z0, s(z1)) → c3(+'(z0, z1))

Defined Rule Symbols:

sum, +

Defined Pair Symbols:

SUM, +'

Compound Symbols:

c1, c3

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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