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:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(X1, X2)
activate(n__from(X)) → from(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__len(X)) → len(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:

FST(s(X), cons(Y, Z)) → ACTIVATE(X)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(s(X), Y) → ACTIVATE(X)
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__fst(X1, X2)) → FST(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ACTIVATE(n__len(X)) → LEN(X)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(X1, X2)
activate(n__from(X)) → from(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__len(X)) → len(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 1 less node.

(4) Obligation:

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

ACTIVATE(n__fst(X1, X2)) → FST(X1, X2)
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__len(X)) → LEN(X)
LEN(cons(X, Z)) → ACTIVATE(Z)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(X1, X2)
activate(n__from(X)) → from(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__len(X)) → len(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.

ACTIVATE(n__fst(X1, X2)) → FST(X1, X2)
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ACTIVATE(n__len(X)) → LEN(X)
LEN(cons(X, Z)) → ACTIVATE(Z)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  ACTIVATE(x1)
n__fst(x1, x2)  =  n__fst(x1, x2)
FST(x1, x2)  =  FST(x1, x2)
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x2)
n__add(x1, x2)  =  n__add(x1)
ADD(x1, x2)  =  x1
n__len(x1)  =  n__len(x1)
LEN(x1)  =  LEN(x1)

Lexicographic Path Order [LPO].
Precedence:

[ACTIVATE₁, s₁, cons₁] > FST₂
[ACTIVATE₁, s₁, cons₁] > LEN₁

The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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

ADD(s(X), Y) → ACTIVATE(X)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(X1, X2)
activate(n__from(X)) → from(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__len(X)) → len(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 1 less node.

(8) TRUE