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:

from(X) → cons(X, n__from(s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, activate(XS))
from(X) → n__from(X)
activate(n__from(X)) → from(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:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
after(0, z0) → z0
after(s(z0), cons(z1, z2)) → after(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(z0) → z0

Tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__from(z0)) → c4(FROM(z0))

S tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__from(z0)) → c4(FROM(z0))

K tuples:none
Defined Rule Symbols:

from, after, activate

Defined Pair Symbols:

AFTER, ACTIVATE

Compound Symbols:

c3, c4

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

Removed 1 trailing nodes:

ACTIVATE(n__from(z0)) → c4(FROM(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
after(0, z0) → z0
after(s(z0), cons(z1, z2)) → after(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(z0) → z0

Tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2))

S tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2))

K tuples:none
Defined Rule Symbols:

from, after, activate

Defined Pair Symbols:

AFTER

Compound Symbols:

c3

(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.

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2))

We considered the (Usable) Rules:

activate(n__from(z0)) → from(z0)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)

And the Tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2))

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

POL(ACTIVATE(x₁)) = 0   
POL(AFTER(x₁, x₂)) = [2]x₁   
POL(activate(x₁)) = 0   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(cons(x₁, x₂)) = 0   
POL(from(x₁)) = 0   
POL(n__from(x₁)) = 0   
POL(s(x₁)) = [1] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
after(0, z0) → z0
after(s(z0), cons(z1, z2)) → after(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(z0) → z0

Tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2))

S tuples:none
K tuples:

AFTER(s(z0), cons(z1, z2)) → c3(AFTER(z0, activate(z2)), ACTIVATE(z2))

Defined Rule Symbols:

from, after, activate

Defined Pair Symbols:

AFTER

Compound Symbols:

c3

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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