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 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtLeafRemovalProof (ComplexityIfPolyImplication, 0 ms)
↳8 CdtProblem
↳9 CdtUsableRulesProof (⇔, 0 ms)
↳10 CdtProblem
↳11 CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))), 123 ms)
↳12 CdtProblem
↳13 CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^2))), 53 ms)
↳14 CdtProblem
↳15 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳16 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 Cpx (relative) TRS 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:

FROM(z0) → c
FROM(z0) → c1
HEAD(cons(z0, z1)) → c2
2ND(cons(z0, z1)) → c3(HEAD(activate(z1)), ACTIVATE(z1))
TAKE(0, z0) → c4
TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
TAKE(z0, z1) → c6
SEL(0, cons(z0, z1)) → c7
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))
ACTIVATE(z0) → c11
S tuples:

FROM(z0) → c
FROM(z0) → c1
HEAD(cons(z0, z1)) → c2
2ND(cons(z0, z1)) → c3(HEAD(activate(z1)), ACTIVATE(z1))
TAKE(0, z0) → c4
TAKE(s(z0), cons(z1, z2)) → c5(ACTIVATE(z2))
TAKE(z0, z1) → c6
SEL(0, cons(z0, z1)) → c7
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))
ACTIVATE(z0) → c11
K tuples:none
Defined Rule Symbols:

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

Defined Pair Symbols:

FROM, HEAD, 2ND, TAKE, SEL, ACTIVATE

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11

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

Removed 8 trailing nodes:

ACTIVATE(n__from(z0)) → c9(FROM(z0))
FROM(z0) → c
TAKE(0, z0) → c4
FROM(z0) → c1
HEAD(cons(z0, z1)) → c2
ACTIVATE(z0) → c11
TAKE(z0, z1) → c6
SEL(0, cons(z0, z1)) → c7

(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) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

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))
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:none
Defined Rule Symbols:

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

Defined Pair Symbols:

TAKE, SEL, ACTIVATE

Compound Symbols:

c5, c8, c10

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

head(cons(z0, z1)) → z0
2nd(cons(z0, z1)) → head(activate(z1))
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

activate(n__from(z0)) → from(z0)
activate(n__take(z0, z1)) → take(z0, z1)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(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)
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))
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:none
Defined Rule Symbols:

activate, from, take

Defined Pair Symbols:

TAKE, SEL, ACTIVATE

Compound Symbols:

c5, c8, c10

(11) CdtRuleRemovalProof (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:none
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))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ACTIVATE(x1)) = 0   
POL(SEL(x1, x2)) = x1   
POL(TAKE(x1, x2)) = 0   
POL(activate(x1)) = [4]x1   
POL(c10(x1)) = x1   
POL(c5(x1)) = x1   
POL(c8(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = [3] + x1   
POL(from(x1)) = [4] + [2]x1   
POL(n__from(x1)) = [3] + x1   
POL(n__take(x1, x2)) = [1] + x1 + x2   
POL(nil) = [3]   
POL(s(x1)) = [1] + x1   
POL(take(x1, x2)) = [3] + [2]x1 + [2]x2   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

activate(n__from(z0)) → from(z0)
activate(n__take(z0, z1)) → take(z0, z1)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(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)
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))
S tuples:

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

SEL(s(z0), cons(z1, z2)) → c8(SEL(z0, activate(z2)), ACTIVATE(z2))
Defined Rule Symbols:

activate, from, take

Defined Pair Symbols:

TAKE, SEL, ACTIVATE

Compound Symbols:

c5, c8, c10

(13) CdtRuleRemovalProof (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(z0) → z0
activate(n__from(z0)) → from(z0)
take(s(z0), cons(z1, z2)) → cons(z1, n__take(z0, activate(z2)))
from(z0) → n__from(z0)
take(z0, z1) → n__take(z0, z1)
from(z0) → cons(z0, n__from(s(z0)))
take(0, z0) → nil
activate(n__take(z0, z1)) → take(z0, z1)
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))
The order we found is given by the following interpretation:
Polynomial interpretation :

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

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

activate(n__from(z0)) → from(z0)
activate(n__take(z0, z1)) → take(z0, z1)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(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)
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))
S tuples:none
K tuples:

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:

activate, from, take

Defined Pair Symbols:

TAKE, SEL, ACTIVATE

Compound Symbols:

c5, c8, c10

(15) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(16) BOUNDS(O(1), O(1))