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, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

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, n__f(s(0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
activate(n__f(z0)) → f(z0)
activate(z0) → z0

Tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0)))
ACTIVATE(n__f(z0)) → c4(F(z0))

S tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0)))
ACTIVATE(n__f(z0)) → c4(F(z0))

K tuples:none
Defined Rule Symbols:

f, p, activate

Defined Pair Symbols:

F, ACTIVATE

Compound Symbols:

c1, c4

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

Removed 1 of 2 dangling nodes:

ACTIVATE(n__f(z0)) → c4(F(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
activate(n__f(z0)) → f(z0)
activate(z0) → z0

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, activate

Defined Pair Symbols:

F

Compound Symbols:

c1

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
activate(n__f(z0)) → f(z0)
activate(z0) → z0

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, activate

Defined Pair Symbols:

F

Compound Symbols:

c1

(7) 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   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
activate(n__f(z0)) → f(z0)
activate(z0) → z0

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, activate

Defined Pair Symbols:

F

Compound Symbols:

c1

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(10) BOUNDS(O(1), O(1))