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

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtKnowledgeProof (⇔, 0 ms)
↳6 CdtProblem
↳7 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))), 226 ms)
↳8 CdtProblem
↳9 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))), 195 ms)
↳10 CdtProblem
↳11 SIsEmptyProof (⇔, 0 ms)
↳12 BOUNDS(O(1), O(1))

(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) CdtKnowledgeProof (EQUIVALENT transformation)

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

2ND(cons(z0, z1)) → c3(HEAD(activate(z1)), ACTIVATE(z1))

(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:

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:

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(HEAD(activate(z1)), ACTIVATE(z1))
Defined Rule Symbols:

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

Defined Pair Symbols:

2ND, TAKE, SEL, ACTIVATE

Compound Symbols:

c3, c5, c8, c10

(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.

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:

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

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

(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:

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:

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(HEAD(activate(z1)), 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:

2ND, TAKE, SEL, ACTIVATE

Compound Symbols:

c3, c5, c8, c10

(9) 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.

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

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

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

(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:

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:none
K tuples:

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

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

Defined Pair Symbols:

2ND, TAKE, SEL, ACTIVATE

Compound Symbols:

c3, c5, c8, c10

(11) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(12) BOUNDS(O(1), O(1))