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 with status [RPO].
Precedence:

fst₂ > nil > 0 > s
fst₂ > cons₁ > s
from₁ > cons₁ > s
add₂ > s
len₁ > 0 > s

Status:

add₂: multiset
from₁: multiset
cons₁: multiset
fst₂: multiset
s: multiset
0: multiset
nil: multiset
len₁: multiset

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