deffunc H₁( Element of 3 -tuples_on F₁()) -> Element of F₁() = F₂(($1 . 1),($1 . 2),($1 . 3));
consider f being Function of (3 -tuples_on F₁()),F₁() such that
A1: for a being Element of 3 -tuples_on F₁() holds f . a = H₁(a) from FUNCT_2:sch 4();

hereby :: thesis: for f1, f2 being Function of (3 -tuples_on F₁()),F₁() st ( for x, y, z being Element of F₁() holds f1 . <*x,y,z*> = F₂(x,y,z) ) & ( for x, y, z being Element of F₁() holds f2 . <*x,y,z*> = F₂(x,y,z) ) holds
f1 = f2

take f = f; :: thesis: for x, y, z being Element of F₁() holds f . <*x,y,z*> = F₂(x,y,z)
let x, y, z be Element of F₁(); :: thesis: f . <*x,y,z*> = F₂(x,y,z)
reconsider a = <*x,y,z*> as Element of 3 -tuples_on F₁() by FINSEQ_2:104;
thus f . <*x,y,z*> = F₂((a . 1),(a . 2),(a . 3)) by A1
.= F₂(x,(a . 2),(a . 3))
.= F₂(x,y,(a . 3))
.= F₂(x,y,z) ; :: thesis: verum

end;

let f1, f2 be Function of (3 -tuples_on F₁()),F₁(); :: thesis: ( ( for x, y, z being Element of F₁() holds f1 . <*x,y,z*> = F₂(x,y,z) ) & ( for x, y, z being Element of F₁() holds f2 . <*x,y,z*> = F₂(x,y,z) ) implies f1 = f2 )
assume that
A2: for x, y, z being Element of F₁() holds f1 . <*x,y,z*> = F₂(x,y,z) and
A3: for x, y, z being Element of F₁() holds f2 . <*x,y,z*> = F₂(x,y,z) ; :: thesis: f1 = f2

now :: thesis: for a being Element of 3 -tuples_on F₁() holds f1 . a = f2 . a

let a be Element of 3 -tuples_on F₁(); :: thesis: f1 . a = f2 . a
consider x, y, z being Element of F₁() such that
A4: a = <*x,y,z*> by FINSEQ_2:103;
thus f1 . a = F₂(x,y,z) by A2, A4
.= f2 . a by A3, A4 ; :: thesis: verum

end;

hence f1 = f2 by FUNCT_2:63; :: thesis: verum