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(s(x), y) → f(x, g(x, y))
f(0, y) → y
g(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(s(z0), z1) → f(z0, g(z0, z1))
f(0, z0) → z0
g(z0, z1) → z1

Tuples:

F(s(z0), z1) → c(F(z0, g(z0, z1)), G(z0, z1))

S tuples:

F(s(z0), z1) → c(F(z0, g(z0, z1)), G(z0, z1))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F

Compound Symbols:

c

(3) CdtGraphRemoveTrailingProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(s(z0), z1) → c(F(z0, g(z0, z1)))

S tuples:

F(s(z0), z1) → c(F(z0, g(z0, z1)))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F

Compound Symbols:

c

(5) 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) → c(F(z0, g(z0, z1)))

We considered the (Usable) Rules:

g(z0, z1) → z1

And the Tuples:

F(s(z0), z1) → c(F(z0, g(z0, z1)))

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

POL(F(x₁, x₂)) = [4]x₁ + [3]x₂   
POL(c(x₁)) = x₁   
POL(g(x₁, x₂)) = [1] + x₂   
POL(s(x₁)) = [4] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(s(z0), z1) → c(F(z0, g(z0, z1)))

S tuples:none
K tuples:

F(s(z0), z1) → c(F(z0, g(z0, z1)))

Defined Rule Symbols:

f, g

Defined Pair Symbols:

F

Compound Symbols:

c

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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