theorem Th28:
  ex n st square-uparrow n c= [:(NAT \ Segm i),(NAT \ Segm j):]
  proof
    reconsider n = max(i,j) as Element of NAT by ORDINAL1:def 12;
    take n;
    let x be object;
    assume
A1: x in square-uparrow n;
    then reconsider y = x as Element of [:NAT,NAT:];
    consider n1,n2 be Nat such that
A2: y`1 = n1 and
A3: y`2 = n2 and
A4: n <= n1 and
A5: n <= n2 by A1,Def3;
A6: y is pair by Th4;
    i <= n by XXREAL_0:25; then
A7: not n1 in Segm i by A4,XXREAL_0:2,NAT_1:44;
    n1 in NAT by ORDINAL1:def 12; then
A8: n1 in NAT \ Segm i by A7,XBOOLE_0:def 5;
    j <= n by XXREAL_0:25; then
A9: not n2 in Segm j by A5,XXREAL_0:2,NAT_1:44;
    n2 in NAT by ORDINAL1:def 12;
    then n2 in NAT \ Segm j by A9,XBOOLE_0:def 5;
    hence thesis by A2,A3,A6,A8,ZFMISC_1:def 2;
  end;
