### (0) Obligation:

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

from(X) → cons(X, n__from(s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
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 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)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:

FROM(z0) → c
FROM(z0) → c1
SEL(0, cons(z0, z1)) → c2
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__from(z0)) → c4(FROM(z0))
ACTIVATE(z0) → c5
S tuples:

FROM(z0) → c
FROM(z0) → c1
SEL(0, cons(z0, z1)) → c2
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__from(z0)) → c4(FROM(z0))
ACTIVATE(z0) → c5
K tuples:none
Defined Rule Symbols:

from, sel, activate

Defined Pair Symbols:

FROM, SEL, ACTIVATE

Compound Symbols:

c, c1, c2, c3, c4, c5

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

Removed 5 trailing nodes:

FROM(z0) → c
FROM(z0) → c1
ACTIVATE(z0) → c5
ACTIVATE(n__from(z0)) → c4(FROM(z0))
SEL(0, cons(z0, z1)) → c2

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2))
S tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2))
K tuples:none
Defined Rule Symbols:

from, sel, activate

Defined Pair Symbols:

SEL

Compound Symbols:

c3

### (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)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
S tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
K tuples:none
Defined Rule Symbols:

from, sel, activate

Defined Pair Symbols:

SEL

Compound Symbols:

c3

### (7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
S tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
K tuples:none
Defined Rule Symbols:

activate, from

Defined Pair Symbols:

SEL

Compound Symbols:

c3

### (9) 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)) → c3(SEL(z0, activate(z2)))
We considered the (Usable) Rules:none
And the Tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(SEL(x1, x2)) = x1
POL(activate(x1)) = 0
POL(c3(x1)) = x1
POL(cons(x1, x2)) = 0
POL(from(x1)) = [2] + [3]x1
POL(n__from(x1)) = 0
POL(s(x1)) = [1] + x1

### (10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
S tuples:none
K tuples:

SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
Defined Rule Symbols:

activate, from

Defined Pair Symbols:

SEL

Compound Symbols:

c3

### (11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty