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:

f(0) → s(0)
f(s(0)) → s(0)
f(s(s(x))) → f(f(s(x)))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → s(0)
f(s(0)) → s(0)
f(s(s(z0))) → f(f(s(z0)))

Tuples:

F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

S tuples:

F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2

(3) 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.

F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

We considered the (Usable) Rules:

f(s(0)) → s(0)
f(s(s(z0))) → f(f(s(z0)))
f(0) → s(0)

And the Tuples:

F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

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

POL(0) = 0   
POL(F(x₁)) = [2]x₁²   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(f(x₁)) = [2]   
POL(s(x₁)) = [2] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → s(0)
f(s(0)) → s(0)
f(s(s(z0))) → f(f(s(z0)))

Tuples:

F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

S tuples:none
K tuples:

F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2

(5) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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