theorem Th29:
  n = max(i,j) implies
  square-uparrow n c= square-uparrow i /\ square-uparrow j
  proof
    assume
A1: n = max(i,j);
    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
A6: n <= n2 by A2,Def3;
    i <= n & j <= n by A1,XXREAL_0:25;
    then i <= n1 & j <= n1 & i <= n2 & j <= n2 by A5,A6,XXREAL_0:2;
    then y in square-uparrow i & y in square-uparrow j by A3,A4,Def3;
    hence thesis by XBOOLE_0:def 4;
  end;
