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, cons(Y)) → cons(Y)
from(X) → cons(X)
add(0, X) → X
add(s, Y) → s
len(nil) → 0
len(cons(X)) → s

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive Path Order [RPO].
Precedence:

from₁ > [fst₂, cons₁] > [0, nil]
from₁ > [fst₂, cons₁] > [s, add₂]
len₁ > [0, nil]
len₁ > [s, add₂]

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

fst(0, Z) → nil
fst(s, cons(Y)) → cons(Y)
from(X) → cons(X)
add(0, X) → X
add(s, Y) → s
len(nil) → 0
len(cons(X)) → s

(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