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:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

AND(true, X) → ACTIVATE(X)
IF(true, X, Y) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
ADD(0, X) → ACTIVATE(X)
ADD(s(X), Y) → S(n__add(activate(X), activate(Y)))
ADD(s(X), Y) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(Y)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
FROM(X) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), X2)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__s(X)) → S(X)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), X2)
ADD(0, X) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → FROM(X)
FROM(X) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(s(X), Y) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(Y)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADD(0, X) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
FROM(X) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(s(X), Y) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(ACTIVATE(x₁)) = 5/4 + x₁   
POL(n__add(x₁, x₂)) = 1/2 + (4)x₁ + (7/4)x₂   
POL(ADD(x₁, x₂)) = 7/4 + (4)x₁ + (7/4)x₂   
POL(activate(x₁)) = x₁   
POL(0) = 1/2   
POL(n__first(x₁, x₂)) = 11/4 + (4)x₁ + (4)x₂   
POL(FIRST(x₁, x₂)) = 4 + (4)x₁ + (4)x₂   
POL(s(x₁)) = 1/2 + (1/4)x₁   
POL(cons(x₁, x₂)) = (2)x₁ + (1/4)x₂   
POL(n__from(x₁)) = 1/2 + (9/4)x₁   
POL(FROM(x₁)) = 7/4 + (7/4)x₁   
POL(from(x₁)) = 1/2 + (9/4)x₁   
POL(first(x₁, x₂)) = 11/4 + (4)x₁ + (4)x₂   
POL(n__s(x₁)) = 1/2 + (1/4)x₁   
POL(add(x₁, x₂)) = 1/2 + (4)x₁ + (7/4)x₂   
POL(nil) = 3   

The value of delta used in the strict ordering is 1/2.
The following usable rules [FROCOS05] were oriented:

activate(n__from(X)) → from(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X
activate(n__s(X)) → s(X)
from(X) → n__from(X)
first(X1, X2) → n__first(X1, X2)
activate(n__add(X1, X2)) → add(activate(X1), X2)
add(0, X) → activate(X)
s(X) → n__s(X)
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), X2)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__from(X)) → FROM(X)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

(8) TRUE