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

(0) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(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(n__s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, n__cons(z2, z3))) → rcons(posrecip(activate(z2)), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, n__cons(z2, z3))) → rcons(negrecip(activate(z2)), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
s(z0) → n__s(z0)
cons(z0, z1) → n__cons(z0, z1)
activate(n__from(z0)) → from(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__cons(z0, z1)) → cons(activate(z0), z1)
activate(z0) → z0

Tuples:

FROM(z0) → c(CONS(z0, n__from(n__s(z0))))
2NDSPOS(s(z0), cons(z1, n__cons(z2, z3))) → c3(ACTIVATE(z2), 2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, n__cons(z2, z3))) → c5(ACTIVATE(z2), 2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c6(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c8(S(plus(z0, z1)), PLUS(z0, z1))
TIMES(s(z0), z1) → c10(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
SQUARE(z0) → c11(TIMES(z0, z0))
ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c15(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__cons(z0, z1)) → c16(CONS(activate(z0), z1), ACTIVATE(z0))

S tuples:

FROM(z0) → c(CONS(z0, n__from(n__s(z0))))
2NDSPOS(s(z0), cons(z1, n__cons(z2, z3))) → c3(ACTIVATE(z2), 2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, n__cons(z2, z3))) → c5(ACTIVATE(z2), 2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PI(z0) → c6(2NDSPOS(z0, from(0)), FROM(0))
PLUS(s(z0), z1) → c8(S(plus(z0, z1)), PLUS(z0, z1))
TIMES(s(z0), z1) → c10(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
SQUARE(z0) → c11(TIMES(z0, z0))
ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c15(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__cons(z0, z1)) → c16(CONS(activate(z0), z1), ACTIVATE(z0))

K tuples:none
Defined Rule Symbols:

from, 2ndspos, 2ndsneg, pi, plus, times, square, s, cons, activate

Defined Pair Symbols:

FROM, 2NDSPOS, 2NDSNEG, PI, PLUS, TIMES, SQUARE, ACTIVATE

Compound Symbols:

c, c3, c5, c6, c8, c10, c11, c14, c15, c16

(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

2NDSPOS(s(z0), cons(z1, n__cons(z2, z3))) → c3(ACTIVATE(z2), 2NDSNEG(z0, activate(z3)), ACTIVATE(z3))
2NDSNEG(s(z0), cons(z1, n__cons(z2, z3))) → c5(ACTIVATE(z2), 2NDSPOS(z0, activate(z3)), ACTIVATE(z3))
PLUS(s(z0), z1) → c8(S(plus(z0, z1)), PLUS(z0, z1))
TIMES(s(z0), z1) → c10(PLUS(z1, times(z0, z1)), TIMES(z0, z1))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, n__cons(z2, z3))) → rcons(posrecip(activate(z2)), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, n__cons(z2, z3))) → rcons(negrecip(activate(z2)), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
s(z0) → n__s(z0)
cons(z0, z1) → n__cons(z0, z1)
activate(n__from(z0)) → from(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__cons(z0, z1)) → cons(activate(z0), z1)
activate(z0) → z0

Tuples:

FROM(z0) → c(CONS(z0, n__from(n__s(z0))))
PI(z0) → c6(2NDSPOS(z0, from(0)), FROM(0))
SQUARE(z0) → c11(TIMES(z0, z0))
ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c15(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__cons(z0, z1)) → c16(CONS(activate(z0), z1), ACTIVATE(z0))

S tuples:

FROM(z0) → c(CONS(z0, n__from(n__s(z0))))
PI(z0) → c6(2NDSPOS(z0, from(0)), FROM(0))
SQUARE(z0) → c11(TIMES(z0, z0))
ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c15(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__cons(z0, z1)) → c16(CONS(activate(z0), z1), ACTIVATE(z0))

K tuples:none
Defined Rule Symbols:

from, 2ndspos, 2ndsneg, pi, plus, times, square, s, cons, activate

Defined Pair Symbols:

FROM, PI, SQUARE, ACTIVATE

Compound Symbols:

c, c6, c11, c14, c15, c16

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

Removed 3 trailing nodes:

PI(z0) → c6(2NDSPOS(z0, from(0)), FROM(0))
FROM(z0) → c(CONS(z0, n__from(n__s(z0))))
SQUARE(z0) → c11(TIMES(z0, z0))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, n__cons(z2, z3))) → rcons(posrecip(activate(z2)), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, n__cons(z2, z3))) → rcons(negrecip(activate(z2)), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
s(z0) → n__s(z0)
cons(z0, z1) → n__cons(z0, z1)
activate(n__from(z0)) → from(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__cons(z0, z1)) → cons(activate(z0), z1)
activate(z0) → z0

Tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c15(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__cons(z0, z1)) → c16(CONS(activate(z0), z1), ACTIVATE(z0))

S tuples:

ACTIVATE(n__from(z0)) → c14(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c15(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__cons(z0, z1)) → c16(CONS(activate(z0), z1), ACTIVATE(z0))

K tuples:none
Defined Rule Symbols:

from, 2ndspos, 2ndsneg, pi, plus, times, square, s, cons, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c14, c15, c16

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

Removed 3 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, n__cons(z2, z3))) → rcons(posrecip(activate(z2)), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, n__cons(z2, z3))) → rcons(negrecip(activate(z2)), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
s(z0) → n__s(z0)
cons(z0, z1) → n__cons(z0, z1)
activate(n__from(z0)) → from(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__cons(z0, z1)) → cons(activate(z0), z1)
activate(z0) → z0

Tuples:

ACTIVATE(n__from(z0)) → c14(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0))
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))

S tuples:

ACTIVATE(n__from(z0)) → c14(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0))
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))

K tuples:none
Defined Rule Symbols:

from, 2ndspos, 2ndsneg, pi, plus, times, square, s, cons, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c14, c15, c16

(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.

ACTIVATE(n__from(z0)) → c14(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0))
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))

We considered the (Usable) Rules:none
And the Tuples:

ACTIVATE(n__from(z0)) → c14(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0))
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))

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

POL(ACTIVATE(x₁)) = [4]x₁   
POL(c14(x₁)) = x₁   
POL(c15(x₁)) = x₁   
POL(c16(x₁)) = x₁   
POL(n__cons(x₁, x₂)) = [4] + x₁   
POL(n__from(x₁)) = [2] + x₁   
POL(n__s(x₁)) = [5] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
2ndspos(0, z0) → rnil
2ndspos(s(z0), cons(z1, n__cons(z2, z3))) → rcons(posrecip(activate(z2)), 2ndsneg(z0, activate(z3)))
2ndsneg(0, z0) → rnil
2ndsneg(s(z0), cons(z1, n__cons(z2, z3))) → rcons(negrecip(activate(z2)), 2ndspos(z0, activate(z3)))
pi(z0) → 2ndspos(z0, from(0))
plus(0, z0) → z0
plus(s(z0), z1) → s(plus(z0, z1))
times(0, z0) → 0
times(s(z0), z1) → plus(z1, times(z0, z1))
square(z0) → times(z0, z0)
s(z0) → n__s(z0)
cons(z0, z1) → n__cons(z0, z1)
activate(n__from(z0)) → from(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__cons(z0, z1)) → cons(activate(z0), z1)
activate(z0) → z0

Tuples:

ACTIVATE(n__from(z0)) → c14(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0))
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))

S tuples:none
K tuples:

ACTIVATE(n__from(z0)) → c14(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c15(ACTIVATE(z0))
ACTIVATE(n__cons(z0, z1)) → c16(ACTIVATE(z0))

Defined Rule Symbols:

from, 2ndspos, 2ndsneg, pi, plus, times, square, s, cons, activate

Defined Pair Symbols:

ACTIVATE

Compound Symbols:

c14, c15, c16

(11) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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