let I be non empty set ; :: thesis: for J being RelStr-yielding non-Empty reflexive-yielding ManySortedSet of I st ( for i being Element of I holds J . i is sup-Semilattice ) holds
for f, g being Element of ()
for i being Element of I holds (f "\/" g) . i = (f . i) "\/" (g . i)

let J be RelStr-yielding non-Empty reflexive-yielding ManySortedSet of I; :: thesis: ( ( for i being Element of I holds J . i is sup-Semilattice ) implies for f, g being Element of ()
for i being Element of I holds (f "\/" g) . i = (f . i) "\/" (g . i) )

assume A1: for i being Element of I holds J . i is sup-Semilattice ; :: thesis: for f, g being Element of ()
for i being Element of I holds (f "\/" g) . i = (f . i) "\/" (g . i)

let f, g be Element of (); :: thesis: for i being Element of I holds (f "\/" g) . i = (f . i) "\/" (g . i)
let i be Element of I; :: thesis: (f "\/" g) . i = (f . i) "\/" (g . i)
A2: J . i is sup-Semilattice by A1;
for i being Element of I holds J . i is antisymmetric by A1;
then A3: ( product J is antisymmetric & product J is with_suprema ) by ;
then g <= f "\/" g by YELLOW_0:22;
then A4: g . i <= (f "\/" g) . i by WAYBEL_3:28;
A5: (f "\/" g) . i <= (f . i) "\/" (g . i)
proof
deffunc H1( Element of I) -> Element of the carrier of (J . \$1) = (f . \$1) "\/" (g . \$1);
consider z being ManySortedSet of I such that
A6: for i being Element of I holds z . i = H1(i) from PBOOLE:sch 5();
A7: for i being Element of I holds z . i is Element of (J . i)
proof
let i be Element of I; :: thesis: z . i is Element of (J . i)
z . i = (f . i) "\/" (g . i) by A6;
hence z . i is Element of (J . i) ; :: thesis: verum
end;
dom z = I by PARTFUN1:def 2;
then reconsider z = z as Element of () by ;
for i being Element of I holds
( z . i >= f . i & z . i >= g . i )
proof
let i be Element of I; :: thesis: ( z . i >= f . i & z . i >= g . i )
( J . i is antisymmetric with_suprema RelStr & z . i = (f . i) "\/" (g . i) ) by A1, A6;
hence ( z . i >= f . i & z . i >= g . i ) by YELLOW_0:22; :: thesis: verum
end;
then ( z >= f & z >= g ) by WAYBEL_3:28;
then A8: f "\/" g <= z by ;
for i being Element of I holds (f "\/" g) . i <= (f . i) "\/" (g . i)
proof
let i be Element of I; :: thesis: (f "\/" g) . i <= (f . i) "\/" (g . i)
(f . i) "\/" (g . i) = z . i by A6;
hence (f "\/" g) . i <= (f . i) "\/" (g . i) by ; :: thesis: verum
end;
hence (f "\/" g) . i <= (f . i) "\/" (g . i) ; :: thesis: verum
end;
f <= f "\/" g by ;
then f . i <= (f "\/" g) . i by WAYBEL_3:28;
then (f "\/" g) . i >= (f . i) "\/" (g . i) by ;
hence (f "\/" g) . i = (f . i) "\/" (g . i) by ; :: thesis: verum