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, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0, 1, z0) → f(s(z0), z0, z0)
f(z0, z1, s(z2)) → s(f(0, 1, z2))

Tuples:

F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))

S tuples:

F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, 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(z0, z1, s(z2)) → c1(F(0, 1, z2))

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

F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))

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

POL(0) = [4]   
POL(1) = 0   
POL(F(x₁, x₂, x₃)) = [4]x₃   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(s(x₁)) = [4] + x₁   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0, 1, z0) → f(s(z0), z0, z0)
f(z0, z1, s(z2)) → s(f(0, 1, z2))

Tuples:

F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))

S tuples:

F(0, 1, z0) → c(F(s(z0), z0, z0))

K tuples:

F(z0, z1, s(z2)) → c1(F(0, 1, z2))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

(5) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

F(0, 1, z0) → c(F(s(z0), z0, z0))
F(z0, z1, s(z2)) → c1(F(0, 1, z2))

Now S is empty

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