let X, Y be non empty reflexive RelStr ; ( [:X,Y:] is with_suprema implies ( X is with_suprema & Y is with_suprema ) )
assume A1:
[:X,Y:] is with_suprema
; ( X is with_suprema & Y is with_suprema )
A2:
the carrier of [:X,Y:] = [:the carrier of X,the carrier of Y:]
by Def2;
thus
X is with_suprema
Y is with_suprema proof
let x,
y be
Element of ;
LATTICE3:def 10 ex b1 being Element of the carrier of X st
( x <= b1 & y <= b1 & ( for b2 being Element of the carrier of X holds
( not x <= b2 or not y <= b2 or b1 <= b2 ) ) )
consider a being
Element of ;
A3:
a <= a
;
consider z being
Element of
such that A4:
(
[x,a] <= z &
[y,a] <= z )
and A5:
for
z' being
Element of st
[x,a] <= z' &
[y,a] <= z' holds
z <= z'
by A1, LATTICE3:def 10;
take
z `1
;
( x <= z `1 & y <= z `1 & ( for b1 being Element of the carrier of X holds
( not x <= b1 or not y <= b1 or z `1 <= b1 ) ) )
A6:
z = [(z `1 ),(z `2 )]
by A2, MCART_1:23;
hence
(
x <= z `1 &
y <= z `1 )
by A4, Th11;
for b1 being Element of the carrier of X holds
( not x <= b1 or not y <= b1 or z `1 <= b1 )
let z' be
Element of ;
( not x <= z' or not y <= z' or z `1 <= z' )
assume
(
x <= z' &
y <= z' )
;
z `1 <= z'
then
(
[x,a] <= [z',a] &
[y,a] <= [z',a] )
by A3, Th11;
then
z <= [z',a]
by A5;
hence
z `1 <= z'
by A6, Th11;
verum
end;
consider a being Element of ;
A7:
a <= a
;
let x, y be Element of ; LATTICE3:def 10 ex b1 being Element of the carrier of Y st
( x <= b1 & y <= b1 & ( for b2 being Element of the carrier of Y holds
( not x <= b2 or not y <= b2 or b1 <= b2 ) ) )
consider z being Element of such that
A8:
( [a,x] <= z & [a,y] <= z )
and
A9:
for z' being Element of st [a,x] <= z' & [a,y] <= z' holds
z <= z'
by A1, LATTICE3:def 10;
take
z `2
; ( x <= z `2 & y <= z `2 & ( for b1 being Element of the carrier of Y holds
( not x <= b1 or not y <= b1 or z `2 <= b1 ) ) )
A10:
z = [(z `1 ),(z `2 )]
by A2, MCART_1:23;
hence
( x <= z `2 & y <= z `2 )
by A8, Th11; for b1 being Element of the carrier of Y holds
( not x <= b1 or not y <= b1 or z `2 <= b1 )
let z' be Element of ; ( not x <= z' or not y <= z' or z `2 <= z' )
assume
( x <= z' & y <= z' )
; z `2 <= z'
then
( [a,x] <= [a,z'] & [a,y] <= [a,z'] )
by A7, Th11;
then
z <= [a,z']
by A9;
hence
z `2 <= z'
by A10, Th11; verum