Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

from(X) → cons(X, n__from(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Lexicographic Path Order [LPO].
Precedence:

pi₁ > [2ndspos₂, 2ndsneg₂] > rnil > [s₁, 0, rcons₂]
pi₁ > [2ndspos₂, 2ndsneg₂] > activate₁ > from₁ > cons₂ > n_cons2 > [s₁, 0, rcons₂]
pi₁ > [2ndspos₂, 2ndsneg₂] > activate₁ > from₁ > cons₂ > posrecip₁ > [s₁, 0, rcons₂]
pi₁ > [2ndspos₂, 2ndsneg₂] > activate₁ > from₁ > cons₂ > negrecip₁ > [s₁, 0, rcons₂]
pi₁ > [2ndspos₂, 2ndsneg₂] > activate₁ > from₁ > n_from1 > [s₁, 0, rcons₂]
square₁ > times₂ > plus₂ > [s₁, 0, rcons₂]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

from(X) → cons(X, n__from(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE