theorem
  for X,Y,Z being set holds
    disjoin <*X,Y,Z*> = <*[:X,{1}:],[:Y,{2}:],[:Z,{3}:]*>
proof
  let X,Y,Z be set;
A1: len disjoin <*X,Y,Z*> = len <*X,Y,Z*> by Th1
    .= 3 by FINSEQ_1:45; then
A2: len disjoin <*X,Y,Z*> = len <*[:X,{1}:],[:Y,{2}:],[:Z,{3}:]*>
    by FINSEQ_1:45;
  now
    let k be Nat;
    assume A3: 1 <= k & k <= len disjoin <*X,Y,Z*>;
    then 1 <= k & k <= len <*X,Y,Z*> by Th1;
    then A4: k in dom <*X,Y,Z*> by Th25;
    k = 1+0 or ... or k = 1+2 by A1, A3, NAT_1:62;
    then per cases;
    suppose A5: k = 1;
      thus (disjoin <*X,Y,Z*>).k = [:<*X,Y,Z*>.k,{k}:] by A4, CARD_3:def 3
        .= [:X,{1}:] by A5
        .= <*[:X,{1}:],[:Y,{2}:],[:Z,{3}:]*>.k by A5;
    end;
    suppose A6: k = 2;
      thus (disjoin <*X,Y,Z*>).k = [:<*X,Y,Z*>.k,{k}:] by A4, CARD_3:def 3
        .= [:Y,{2}:] by A6
        .= <*[:X,{1}:],[:Y,{2}:],[:Z,{3}:]*>.k by A6;
    end;
    suppose A7: k = 3;
      thus (disjoin <*X,Y,Z*>).k = [:<*X,Y,Z*>.k,{k}:] by A4, CARD_3:def 3
        .= [:Z,{3}:] by A7
        .= <*[:X,{1}:],[:Y,{2}:],[:Z,{3}:]*>.k by A7;
    end;
  end;
  hence thesis by A2, FINSEQ_1:def 18;
end;
