reserve x,y, X,Y,Z for set,
        D for non empty set,
        n,k for Nat,
        i,i1,i2 for Integer;

theorem Th10:
  for X be finite c=-linear set st X is with_non-empty_elements
    holds card X c= card union X
 proof
  let X be finite c=-linear set;
  defpred P[Nat] means
   for X be finite c=-linear set
    st X is with_non-empty_elements & card union X=$1
  holds card X c=card union X;
  defpred Q[Nat] means
   for n be Nat st n<=$1 holds P[n];
  assume A1: X is with_non-empty_elements;
  A2: for m be Nat st Q[m] holds Q[m+1]
  proof
   let m be Nat such that
    A3: Q[m];
   let n be Nat;
   assume A4: n<=m+1;
   let X be finite c=-linear set such that
    A5: X is with_non-empty_elements and
    A6: card union X=n;
   reconsider u=union X as finite set by A6;
   reconsider Xu=X\{u} as finite c=-linear set;
   per cases by A4,NAT_1:8;
   suppose n<=m;
    hence thesis by A3,A5,A6;
   end;
   suppose A7: n=m+1;
    then X is non empty by A6,ZFMISC_1:2;
    then A8: X=Xu\/{u} by Th9,ZFMISC_1:116;
    then u=union Xu\/union{u} by ZFMISC_1:78
     .=union Xu\/u by ZFMISC_1:25;
    then A9: union Xu c=u by XBOOLE_1:11;
    then reconsider uXu=union Xu as finite set;
    uXu<>u
    proof
     assume A10: uXu=u;
     then Xu is non empty by A6,A7,ZFMISC_1:2;
     then u in Xu by A10,Th9;
     hence thesis by ZFMISC_1:56;
    end;
    then A11: uXu c<u by A9;
    then card uXu<m+1 by A6,A7,CARD_2:48;
    then card uXu<=m by NAT_1:13;
    then Segm card Xu c= Segm card uXu by A3,A5;
    then card Xu<=card uXu by NAT_1:39;
    then card Xu<card u by A11,CARD_2:48,XXREAL_0:2;
    then
  A12: 1+card Xu<=card u by NAT_1:13;
    not u in Xu by ZFMISC_1:56;
    then
A13:   1+card Xu=card X by A8,CARD_2:41;
     Segm(1+card Xu) c= Segm card u by A12,NAT_1:39;
    hence thesis by A13;
   end;
  end;
  A14: Q[0]
  proof
   let n be Nat;
   assume A15: n<=0;
   let X be finite c=-linear set such that
    A16: X is with_non-empty_elements & card union X=n;
    X is empty by A15,A16;
   hence card X c=card union X;
  end;
   for n be Nat holds Q[n] from NAT_1:sch 2(A14,A2);
   hence thesis by A1;
 end;
