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(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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
from(x1)  =  from(x1)
cons(x1, x2)  =  cons(x1, x2)
n__from(x1)  =  n__from(x1)
n__s(x1)  =  n__s(x1)
2ndspos(x1, x2)  =  2ndspos(x1, x2)
0  =  0
rnil  =  rnil
s(x1)  =  s(x1)
n__cons(x1, x2)  =  n__cons(x1, x2)
rcons(x1, x2)  =  rcons(x1, x2)
posrecip(x1)  =  posrecip(x1)
activate(x1)  =  activate(x1)
2ndsneg(x1, x2)  =  2ndsneg(x1, x2)
negrecip(x1)  =  x1
pi(x1)  =  pi(x1)
plus(x1, x2)  =  plus(x1, x2)
times(x1, x2)  =  times(x1, x2)
square(x1)  =  square(x1)

Recursive path order with status [RPO].
Quasi-Precedence:

pi₁ > [2ndspos₂, rnil, 2ndsneg₂] > posrecip₁ > [n_s1, s₁]
pi₁ > [2ndspos₂, rnil, 2ndsneg₂] > activate₁ > from₁ > cons₂ > n_cons2 > [n_s1, s₁]
pi₁ > [2ndspos₂, rnil, 2ndsneg₂] > activate₁ > from₁ > cons₂ > rcons₂ > [n_s1, s₁]
pi₁ > [2ndspos₂, rnil, 2ndsneg₂] > activate₁ > from₁ > n_from1 > [n_s1, s₁]
pi₁ > 0 > [n_s1, s₁]
square₁ > times₂ > 0 > [n_s1, s₁]
square₁ > times₂ > plus₂ > [n_s1, s₁]

Status:

from₁: [1]
plus₂: [1,2]
rnil: multiset
n_cons2: multiset
rcons₂: multiset
posrecip₁: multiset
activate₁: [1]
n_s1: multiset
pi₁: multiset
square₁: multiset
0: multiset
2ndsneg₂: [1,2]
cons₂: multiset
2ndspos₂: [1,2]
n_from1: multiset
times₂: [1,2]
s₁: multiset

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(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)
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

(2) Obligation:

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

s(X) → n__s(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(n__s(x₁)) = 1 + x₁   
POL(s(x₁)) = 2 + 2·x₁   

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

s(X) → n__s(X)

(4) Obligation:

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

(5) RisEmptyProof (EQUIVALENT transformation)

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

(6) TRUE

(7) RisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE

(9) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(n__s(x₁)) = x₁   
POL(s(x₁)) = 1 + x₁   

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

s(X) → n__s(X)

(10) Obligation:

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