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:

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

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))

S tuples:

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1

(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.

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))

We considered the (Usable) Rules:

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

And the Tuples:

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))

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

POL(0) = 0   
POL(F(x₁, x₂)) = [3] + [4]x₁   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(f(x₁, x₂)) = 0   
POL(s(x₁)) = [4] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))

S tuples:none
K tuples:

F(s(z0), z1) → c1(F(f(z0, z1), z1), F(z0, z1))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c1

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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