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:

from(X) → cons(X, n__from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
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)
head(cons(z0, z1)) → z0
2nd(cons(z0, z1)) → head(activate(z1))
take(0, z0) → nil
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2)))
take(z0, z1) → n__take(z0, z1)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(n__take(z0, z1)) → take(z0, z1)
activate(z0) → z0

Tuples:

2ND(cons(z0, z1)) → c3(HEAD(activate(z1)), ACTIVATE(z1))
TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__from(z0)) → c9(FROM(z0))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))

S tuples:

2ND(cons(z0, z1)) → c3(HEAD(activate(z1)), ACTIVATE(z1))
TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__from(z0)) → c9(FROM(z0))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))

K tuples:none
Defined Rule Symbols:

from, head, 2nd, take, sel, activate

Defined Pair Symbols:

2ND, TAKE, SEL, ACTIVATE

Compound Symbols:

c3, c5, c8, c9, c10

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

Removed 1 trailing nodes:

ACTIVATE(n__from(z0)) → c9(FROM(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
head(cons(z0, z1)) → z0
2nd(cons(z0, z1)) → head(activate(z1))
take(0, z0) → nil
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2)))
take(z0, z1) → n__take(z0, z1)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(n__take(z0, z1)) → take(z0, z1)
activate(z0) → z0

Tuples:

2ND(cons(z0, z1)) → c3(HEAD(activate(z1)), ACTIVATE(z1))
TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))

S tuples:

2ND(cons(z0, z1)) → c3(HEAD(activate(z1)), ACTIVATE(z1))
TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))

K tuples:none
Defined Rule Symbols:

from, head, 2nd, take, sel, activate

Defined Pair Symbols:

2ND, TAKE, SEL, ACTIVATE

Compound Symbols:

c3, c5, c8, c10

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

Removed 1 trailing tuple parts

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
head(cons(z0, z1)) → z0
2nd(cons(z0, z1)) → head(activate(z1))
take(0, z0) → nil
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2)))
take(z0, z1) → n__take(z0, z1)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(n__take(z0, z1)) → take(z0, z1)
activate(z0) → z0

Tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))
2ND(cons(z0, z1)) → c3(ACTIVATE(z1))

S tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))
2ND(cons(z0, z1)) → c3(ACTIVATE(z1))

K tuples:none
Defined Rule Symbols:

from, head, 2nd, take, sel, activate

Defined Pair Symbols:

TAKE, SEL, ACTIVATE, 2ND

Compound Symbols:

c5, c8, c10, c3

(7) CdtKnowledgeProof (EQUIVALENT transformation)

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

2ND(cons(z0, z1)) → c3(ACTIVATE(z1))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
head(cons(z0, z1)) → z0
2nd(cons(z0, z1)) → head(activate(z1))
take(0, z0) → nil
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2)))
take(z0, z1) → n__take(z0, z1)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(n__take(z0, z1)) → take(z0, z1)
activate(z0) → z0

Tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))
2ND(cons(z0, z1)) → c3(ACTIVATE(z1))

S tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))

K tuples:

2ND(cons(z0, z1)) → c3(ACTIVATE(z1))

Defined Rule Symbols:

from, head, 2nd, take, sel, activate

Defined Pair Symbols:

TAKE, SEL, ACTIVATE, 2ND

Compound Symbols:

c5, c8, c10, c3

(9) 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.

SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))

We considered the (Usable) Rules:

activate(n__from(z0)) → from(z0)
activate(n__take(z0, z1)) → take(z0, z1)
activate(z0) → z0
take(0, z0) → nil
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2)))
take(z0, z1) → n__take(z0, z1)
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)

And the Tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))
2ND(cons(z0, z1)) → c3(ACTIVATE(z1))

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

POL(0) = 0   
POL(2ND(x₁)) = [4] + [3]x₁   
POL(ACTIVATE(x₁)) = 0   
POL(SEL(x₁, x₂)) = [4]x₁   
POL(TAKE(x₁, x₂)) = 0   
POL(activate(x₁)) = [2]x₁   
POL(c10(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c8(x₁, x₂)) = x₁ + x₂   
POL(cons(x₁, x₂)) = [3]   
POL(from(x₁)) = [4] + [2]x₁   
POL(n__from(x₁)) = [3] + x₁   
POL(n__take(x₁, x₂)) = [5] + x₁   
POL(nil) = [5]   
POL(s(x₁)) = [4] + x₁   
POL(take(x₁, x₂)) = [5] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
head(cons(z0, z1)) → z0
2nd(cons(z0, z1)) → head(activate(z1))
take(0, z0) → nil
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2)))
take(z0, z1) → n__take(z0, z1)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(n__take(z0, z1)) → take(z0, z1)
activate(z0) → z0

Tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))
2ND(cons(z0, z1)) → c3(ACTIVATE(z1))

S tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))

K tuples:

2ND(cons(z0, z1)) → c3(ACTIVATE(z1))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))

Defined Rule Symbols:

from, head, 2nd, take, sel, activate

Defined Pair Symbols:

TAKE, SEL, ACTIVATE, 2ND

Compound Symbols:

c5, c8, c10, c3

(11) 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__take(z0, z1)) → c10(TAKE(z0, z1))

We considered the (Usable) Rules:

activate(n__from(z0)) → from(z0)
activate(n__take(z0, z1)) → take(z0, z1)
activate(z0) → z0
take(0, z0) → nil
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2)))
take(z0, z1) → n__take(z0, z1)
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)

And the Tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))
2ND(cons(z0, z1)) → c3(ACTIVATE(z1))

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

POL(0) = 0   
POL(2ND(x₁)) = [3] + x₁ + [3]x₁²   
POL(ACTIVATE(x₁)) = x₁   
POL(SEL(x₁, x₂)) = [2]x₁·x₂   
POL(TAKE(x₁, x₂)) = x₂   
POL(activate(x₁)) = x₁   
POL(c10(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c8(x₁, x₂)) = x₁ + x₂   
POL(cons(x₁, x₂)) = x₂   
POL(from(x₁)) = 0   
POL(n__from(x₁)) = 0   
POL(n__take(x₁, x₂)) = [1] + x₂   
POL(nil) = [1]   
POL(s(x₁)) = [1] + x₁   
POL(take(x₁, x₂)) = [1] + x₂   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
head(cons(z0, z1)) → z0
2nd(cons(z0, z1)) → head(activate(z1))
take(0, z0) → nil
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2)))
take(z0, z1) → n__take(z0, z1)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(n__take(z0, z1)) → take(z0, z1)
activate(z0) → z0

Tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))
2ND(cons(z0, z1)) → c3(ACTIVATE(z1))

S tuples:

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))

K tuples:

2ND(cons(z0, z1)) → c3(ACTIVATE(z1))
SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))

Defined Rule Symbols:

from, head, 2nd, take, sel, activate

Defined Pair Symbols:

TAKE, SEL, ACTIVATE, 2ND

Compound Symbols:

c5, c8, c10, c3

(13) CdtKnowledgeProof (EQUIVALENT transformation)

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

TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
ACTIVATE(n__take(z0, z1)) → c10(TAKE(z0, z1))

Now S is empty

(14) BOUNDS(O(1), O(1))