theorem
  m = n - 1 implies square-uparrow n c= [:NAT,NAT:] \ [:Seg m,Seg m:]
  proof
    assume
A1: m = n - 1;
    let x be object;
    assume
A2: x in square-uparrow n;
    then reconsider y = x as Element of [:NAT,NAT:];
    consider n1,n2 be Nat such that
A3: n1 = y`1 and
A4: n2 = y`2 and
A5: n <= n1 and
    n <= n2 by A2,Def3;
    not x in [:Seg m,Seg m:]
    proof
      assume x in [:Seg m,Seg m:];
      then ex x1,x2 be object st x1 in Seg m & x2 in Seg m &
        x = [x1,x2] by ZFMISC_1:def 2;
      then n1 <= m & n2 <= m by A3,A4,FINSEQ_1:1;
      then n <= m by A5,XXREAL_0:2;
      then n - n <= n - 1 - n by A1,XREAL_1:9;
      then 0 <= -1;
      hence thesis;
    end;
    hence thesis by A2,XBOOLE_0:def 5;
  end;
