let L be lower-bounded with_suprema Poset; for X being non empty Subset of (InclPoset (Aux L)) holds meet X is auxiliary Relation of L
let X be non empty Subset of (InclPoset (Aux L)); meet X is auxiliary Relation of L
X c= the carrier of (InclPoset (Aux L))
;
then A1:
X c= Aux L
by YELLOW_1:1;
set a = the Element of X;
the Element of X in X
;
then
the Element of X is auxiliary Relation of L
by A1, Def8;
then reconsider ab = meet X as Relation of L by SETFAM_1:7;
A2:
ab is auxiliary(i)
A6:
ab is auxiliary(ii)
proof
let x,
y,
z,
u be
Element of
L;
WAYBEL_4:def 4 ( u <= x & [x,y] in ab & y <= z implies [u,z] in ab )
assume that A7:
u <= x
and A8:
[x,y] in ab
and A9:
y <= z
;
[u,z] in ab
for
Y being
set st
Y in X holds
[u,z] in Y
proof
let Y be
set ;
( Y in X implies [u,z] in Y )
assume A10:
Y in X
;
[u,z] in Y
then A11:
Y is
auxiliary Relation of
L
by A1, Def8;
[x,y] in Y
by A8, A10, SETFAM_1:def 1;
hence
[u,z] in Y
by A7, A9, A11, Def4;
verum
end;
hence
[u,z] in ab
by SETFAM_1:def 1;
verum
end;
A12:
ab is auxiliary(iii)
proof
let x,
y,
z be
Element of
L;
WAYBEL_4:def 5 ( [x,z] in ab & [y,z] in ab implies [(x "\/" y),z] in ab )
assume that A13:
[x,z] in ab
and A14:
[y,z] in ab
;
[(x "\/" y),z] in ab
for
Y being
set st
Y in X holds
[(x "\/" y),z] in Y
proof
let Y be
set ;
( Y in X implies [(x "\/" y),z] in Y )
assume A15:
Y in X
;
[(x "\/" y),z] in Y
then A16:
Y is
auxiliary Relation of
L
by A1, Def8;
A17:
[x,z] in Y
by A13, A15, SETFAM_1:def 1;
[y,z] in Y
by A14, A15, SETFAM_1:def 1;
hence
[(x "\/" y),z] in Y
by A16, A17, Def5;
verum
end;
hence
[(x "\/" y),z] in ab
by SETFAM_1:def 1;
verum
end;
ab is auxiliary(iv)
hence
meet X is auxiliary Relation of L
by A2, A6, A12; verum