A1: ex B being set st
( B in S & {} c= B & M . B = 0. )

consider B being set such that
A2: ( B = {} & B in S ) by PROB_1:4;
take B ; :: thesis:
thus ( B in S & {} c= B & M . B = 0. ) by A2, VALUED_0:def 19; :: thesis:

end;

A3: {} is Subset of X by XBOOLE_1:2;
A4: for A being set st A = {} holds
ex B being set st
( B in S & ex C being thin of M st A = B \/ C )

reconsider C = {} as thin of M by A3, A1, Def2;
let A be set ; :: thesis:
consider B being set such that
A5: B = {} and
A6: B in S by PROB_1:4;
assume A = {} ; :: thesis:
then A = B \/ C by A5;
hence ex B being set st
( B in S & ex C being thin of M st A = B \/ C ) by A6; :: thesis:

end;

defpred S₁[ set ] means for A being set st A = $1 holds
ex B being set st
( B in S & ex C being thin of M st A = B \/ C );
consider D being set such that
A7: for y being set holds
( y in D iff ( y in bool X & S₁[y] ) ) from XFAMILY:sch 1();
A8: for A being set holds
( A in D iff ex B being set st
( B in S & ex C being thin of M st A = B \/ C ) )

let A be set ; :: thesis:
A9: ( A in D iff ( A in bool X & ( for y being set st y = A holds
ex B being set st
( B in S & ex C being thin of M st y = B \/ C ) ) ) ) by A7;
( ex B being set st
( B in S & ex C being thin of M st A = B \/ C ) implies A in D )

assume A10: ex B being set st
( B in S & ex C being thin of M st A = B \/ C ) ; :: thesis:
then A c= X by XBOOLE_1:8;
hence A in D by A9, A10; :: thesis:

end;

hence ( A in D iff ex B being set st
( B in S & ex C being thin of M st A = B \/ C ) ) by A7; :: thesis:

end;

A11: D c= bool X by A7;
{} c= X ;
then reconsider D = D as non empty Subset-Family of X by A7, A11, A4;
take D ; :: thesis:
thus for A being set holds
( A in D iff ex B being set st
( B in S & ex C being thin of M st A = B \/ C ) ) by A8; :: thesis: