reserve p,q,r for FinSequence;
reserve u,v,x,y,y1,y2,z for object, A,D,X,Y for set;
reserve i,j,k,l,m,n for Nat;
reserve J for Nat;
reserve n for Nat;
reserve x,y,y1,y2,z,a,b for object, X,Y,Z,V1,V2 for set,
  f,g,h,h9,f1,f2 for Function,
  i for Nat,
  P for Permutation of X,
  D,D1,D2,D3 for non empty set,
  d1 for Element of D1,
  d2 for Element of D2,
  d3 for Element of D3;
reserve n for Nat;

theorem
  for X,Y being set holds disjoin <*X,Y*> = <*[:X,{1}:],[:Y,{2}:]*>
proof
  let X,Y be set;
  A1: len disjoin <*X,Y*> = len <*X,Y*> by Th1
    .= 2 by FINSEQ_1:44;
  then A2: len disjoin <*X,Y*> = len <*[:X,{1}:],[:Y,{2}:]*> by FINSEQ_1:44;
  now
    let k be Nat;
    assume A3: 1 <= k & k <= len disjoin <*X,Y*>;
    then 1 <= k & k <= len <*X,Y*> by Th1;
    then A4: k in dom <*X,Y*> by Th25;
    k = 1+0 or ... or k = 1+1 by A1, A3, NAT_1:62;
    then per cases;
    suppose A5: k = 1;
      thus (disjoin <*X,Y*>).k = [:<*X,Y*>.k,{k}:] by A4, CARD_3:def 3
        .= [:X,{1}:] by A5
        .= <*[:X,{1}:],[:Y,{2}:]*>.k by A5;
    end;
    suppose A6: k = 2;
      thus (disjoin <*X,Y*>).k = [:<*X,Y*>.k,{k}:] by A4, CARD_3:def 3
        .= [:Y,{2}:] by A6
        .= <*[:X,{1}:],[:Y,{2}:]*>.k by A6;
    end;
  end;
  hence thesis by A2, FINSEQ_1:def 18;
end;
