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:
Lexicographic path order with status [LPO].
Quasi-Precedence:

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

Status:

add₂: [1,2]
from₁: [1]
cons₁: [1]
fst₂: [2,1]
s: []
0: []
nil: []
len₁: [1]

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

(2) Obligation:

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

from(X) → cons(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Lexicographic path order with status [LPO].
Quasi-Precedence:

from₁ > cons₁

Status:

from₁: [1]
cons₁: [1]

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

from(X) → cons(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