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) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(0)) → 0

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(0)) → 0

Tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0)))

S tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0)))

K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

Compound Symbols:

c1

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

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(0)) → 0

Tuples:

F(s(0)) → c1(F(p(s(0))))

S tuples:

F(s(0)) → c1(F(p(s(0))))

K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

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.

F(s(0)) → c1(F(p(s(0))))

We considered the (Usable) Rules:

p(s(0)) → 0

And the Tuples:

F(s(0)) → c1(F(p(s(0))))

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

POL(0) = [3]   
POL(F(x₁)) = [4]x₁   
POL(c1(x₁)) = x₁   
POL(p(x₁)) = [3]   
POL(s(x₁)) = [5] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(0)) → 0

Tuples:

F(s(0)) → c1(F(p(s(0))))

S tuples:none
K tuples:

F(s(0)) → c1(F(p(s(0))))

Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

Compound Symbols:

c1

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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