The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, activate(X1)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(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:

2nd(cons1(z0, cons(z1, z2))) → z1
2nd(cons(z0, z1)) → 2nd(cons1(z0, activate(z1)))
from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
s(z0) → n__s(z0)
activate(n__from(z0)) → from(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(z0) → z0

Tuples:

2ND(cons(z0, z1)) → c1(2ND(cons1(z0, activate(z1))), ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))

S tuples:

2ND(cons(z0, z1)) → c1(2ND(cons1(z0, activate(z1))), ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))

K tuples:none
Defined Rule Symbols:

2nd, from, s, activate

Defined Pair Symbols:

2ND, ACTIVATE

Compound Symbols:

c1, c5, c6

(3) CdtKnowledgeProof (EQUIVALENT transformation)

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

2ND(cons(z0, z1)) → c1(2ND(cons1(z0, activate(z1))), ACTIVATE(z1))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

2nd(cons1(z0, cons(z1, z2))) → z1
2nd(cons(z0, z1)) → 2nd(cons1(z0, activate(z1)))
from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
s(z0) → n__s(z0)
activate(n__from(z0)) → from(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(z0) → z0

Tuples:

2ND(cons(z0, z1)) → c1(2ND(cons1(z0, activate(z1))), ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))

S tuples:

ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))

K tuples:

2ND(cons(z0, z1)) → c1(2ND(cons1(z0, activate(z1))), ACTIVATE(z1))

Defined Rule Symbols:

2nd, from, s, activate

Defined Pair Symbols:

2ND, ACTIVATE

Compound Symbols:

c1, c5, c6

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))

We considered the (Usable) Rules:

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

And the Tuples:

2ND(cons(z0, z1)) → c1(2ND(cons1(z0, activate(z1))), ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))

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

POL(2ND(x₁)) = x₁ + x₁²   
POL(ACTIVATE(x₁)) = x₁²   
POL(FROM(x₁)) = 0   
POL(S(x₁)) = 0   
POL(activate(x₁)) = 0   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁, x₂)) = x₁ + x₂   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(cons(x₁, x₂)) = x₁ + x₂   
POL(cons1(x₁, x₂)) = 0   
POL(from(x₁)) = 0   
POL(n__from(x₁)) = [1] + x₁   
POL(n__s(x₁)) = [1] + x₁   
POL(s(x₁)) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

2nd(cons1(z0, cons(z1, z2))) → z1
2nd(cons(z0, z1)) → 2nd(cons1(z0, activate(z1)))
from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
s(z0) → n__s(z0)
activate(n__from(z0)) → from(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(z0) → z0

Tuples:

2ND(cons(z0, z1)) → c1(2ND(cons1(z0, activate(z1))), ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))

S tuples:none
K tuples:

2ND(cons(z0, z1)) → c1(2ND(cons1(z0, activate(z1))), ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))

Defined Rule Symbols:

2nd, from, s, activate

Defined Pair Symbols:

2ND, ACTIVATE

Compound Symbols:

c1, c5, c6

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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