theorem
  [i,j] in square-uparrow n implies [i + k,j] in square-uparrow n &
  [i,j + l] in square-uparrow n
  proof
    assume [i,j] in square-uparrow n;
    then consider i1,j1 be Nat such that
A1: i1 = [i,j]`1 and
A2: j1 = [i,j]`2 and
A3: n <= i1 and
A4: n <= j1 by Def3;
    i <= i + k & j <= j + l by NAT_1:11; then
A5: n <= i + k & n <= j + l by A1,A2,A3,A4,XXREAL_0:2;
    i in NAT & i + k in NAT & j in NAT & j + l in NAT by ORDINAL1:def 12;
    then reconsider x = [i + k,j], y = [i,j + l] as Element of [:NAT,NAT:]
      by ZFMISC_1:def 2;
    now
      ex i2,j1 be Nat st i2 = x`1 & j1 = x`2 & n <= i + k & n <= j by A2,A4,A5;
      hence x in square-uparrow n by Def3;
      ex i1,j2 be Nat st i1 = y`1 & j2 = y`2 & n <= i & n <= j + l by A1,A3,A5;
      hence y in square-uparrow n by Def3;
    end;
    hence thesis;
  end;
