let n be Nat; :: thesis: square-downarrow n = [:(Segm n),(Segm n):]
thus square-downarrow n c= [:(Segm n),(Segm n):] :: according to XBOOLE_0:def 10 :: thesis: [:(Segm n),(Segm n):] c= square-downarrow n
proof
let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in square-downarrow n or x in [:(Segm n),(Segm n):] )
assume A2: x in square-downarrow n ; :: thesis: x in [:(Segm n),(Segm n):]
then consider n1, n2 being Nat such that
A3: n1 = x `1 and
A4: n2 = x `2 and
A5: n1 < n and
A6: n2 < n by Def4;
A7: n1 in Segm n by ;
A8: n2 in Segm n by ;
ex x1, x2 being object st
( x1 in NAT & x2 in NAT & x = [x1,x2] ) by ;
then reconsider x = x as pair object ;
x = [n1,n2] by A3, A4;
hence x in [:(Segm n),(Segm n):] by ; :: thesis: verum
end;
let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in [:(Segm n),(Segm n):] or x in square-downarrow n )
assume x in [:(Segm n),(Segm n):] ; :: thesis:
then consider y1, y2 being object such that
A9: y1 in Segm n and
A10: y2 in Segm n and
A11: x = [y1,y2] by ZFMISC_1:def 2;
reconsider y1 = y1, y2 = y2 as Nat by ;
A12: ( y1 = x `1 & y2 = x `2 & y1 < n & y2 < n ) by ;
x is Element of by ;
hence x in square-downarrow n by ; :: thesis: verum