theorem Th30:
  square-downarrow n = [:Segm n,Segm n:]
  proof
    thus square-downarrow n c= [:Segm n,Segm n:]
    proof
      let x be object;
      assume
A2:   x in square-downarrow n;
      then consider n1,n2 be 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 A5,NAT_1:44;
A8:   n2 in Segm n by A6,NAT_1:44;
      ex x1,x2 be object st x1 in NAT & x2 in NAT & x = [x1,x2]
        by A2,ZFMISC_1:def 2;
      then reconsider x as pair object;
      x = [n1,n2] by A3,A4;
      hence thesis by A7,A8,ZFMISC_1:def 2;
    end;
      let x be object;
      assume x in [:Segm n, Segm n:];
      then consider y1, y2 be 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,y2 as Nat by A9,A10;
A12:  y1 = x`1 & y2 = x`2 & y1 < n & y2 < n by A11,A9,A10,NAT_1:44;
      x is Element of [:NAT,NAT:] by A9,A10,A11,ZFMISC_1:def 2;
      hence thesis by A12,Def4;
  end;
