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:

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

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(s(z0), z1) → c2(+'(z0, s(z1)))

S tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(s(z0), z1) → c2(+'(z0, s(z1)))

K tuples:none
Defined Rule Symbols:

+

Defined Pair Symbols:

+'

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.

+'(s(z0), z1) → c1(+'(z0, z1))
+'(s(z0), z1) → c2(+'(z0, s(z1)))

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

+'(s(z0), z1) → c1(+'(z0, z1))
+'(s(z0), z1) → c2(+'(z0, s(z1)))

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

POL(+'(x₁, x₂)) = [4]x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(s(x₁)) = [4] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(s(z0), z1) → c2(+'(z0, s(z1)))

S tuples:none
K tuples:

+'(s(z0), z1) → c1(+'(z0, z1))
+'(s(z0), z1) → c2(+'(z0, s(z1)))

Defined Rule Symbols:

+

Defined Pair Symbols:

+'

Compound Symbols:

c1, c2

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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