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

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

Tuples:

F(s(0)) → c2(F(0))
F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

S tuples:

F(s(0)) → c2(F(0))
F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2, c3, c4

(3) CdtGraphRemoveDanglingProof (ComplexityIfPolyImplication transformation)

Removed 1 of 3 dangling nodes:

F(s(0)) → c2(F(0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

S tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c3, c4

(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(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

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

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

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

POL(+(x₁, x₂)) = [3] + x₁ + x₂   
POL(0) = [1]   
POL(F(x₁)) = [1] + [3]x₁   
POL(c3(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

S tuples:none
K tuples:

F(+(z0, s(0))) → c3(F(z0))
F(+(z0, z1)) → c4(F(z0), F(z1))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c3, c4

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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