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:

g(0) → 0
g(s(x)) → f(g(x))
f(0) → 0

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(0) → 0
g(s(z0)) → f(g(z0))
f(0) → 0

Tuples:

G(s(z0)) → c1(F(g(z0)), G(z0))

S tuples:

G(s(z0)) → c1(F(g(z0)), G(z0))

K tuples:none
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G

Compound Symbols:

c1

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

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(0) → 0
g(s(z0)) → f(g(z0))
f(0) → 0

Tuples:

G(s(z0)) → c1(G(z0))

S tuples:

G(s(z0)) → c1(G(z0))

K tuples:none
Defined Rule Symbols:

g, f

Defined Pair Symbols:

G

Compound Symbols:

c1

(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.

G(s(z0)) → c1(G(z0))

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

G(s(z0)) → c1(G(z0))

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

POL(G(x₁)) = [5]x₁   
POL(c1(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(0) → 0
g(s(z0)) → f(g(z0))
f(0) → 0

Tuples:

G(s(z0)) → c1(G(z0))

S tuples:none
K tuples:

G(s(z0)) → c1(G(z0))

Defined Rule Symbols:

g, f

Defined Pair Symbols:

G

Compound Symbols:

c1

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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