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
A1: for y being set holds
( y in D iff ( y in bool X & S₁[y] ) ) from XBOOLE_0:sch 1();
A2: D c= bool X

proof

let x be set ; :: according to TARSKI:def 3 :: thesis: ( not x in D or x in bool X )
assume x in D ; :: thesis: x in bool X
hence x in bool X by A1; :: thesis: verum

end;

A3: 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 ) )

proof

let A be set ; :: thesis: ( A in D iff ex B being set st
( B in S & ex C being thin of M st A = B \/ C ) )
A4: ( 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 A1;
( ex B being set st
( B in S & ex C being thin of M st A = B \/ C ) implies A in D )

proof

assume A5: ex B being set st
( B in S & ex C being thin of M st A = B \/ C ) ; :: thesis: A in D
then consider B being set such that
A6: ( B in S & ex C being thin of M st A = B \/ C ) ;
A c= X by A6, XBOOLE_1:8;
hence A in D by A4, A5; :: thesis: verum

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 A1; :: thesis: verum

end;

A7: {} is Subset of X by XBOOLE_1:2;
A8: ex B being set st
( B in S & {} c= B & M . B = 0. )

proof

consider B being set such that
A9: ( B = {} & B in S ) by PROB_1:10;
take B ; :: thesis: ( B in S & {} c= B & M . B = 0. )
thus ( B in S & {} c= B & M . B = 0. ) by A9, VALUED_0:def 19; :: thesis: verum

end;

A10: {} c= X by XBOOLE_1:2;
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 )

proof

let A be set ; :: thesis: ( A = {} implies ex B being set st
( B in S & ex C being thin of M st A = B \/ C ) )
assume A11: A = {} ; :: thesis: ex B being set st
( B in S & ex C being thin of M st A = B \/ C )
consider B being set such that
A12: ( B = {} & B in S ) by PROB_1:10;
reconsider C = {} as thin of M by A7, A8, Def3;
A = B \/ C by A11, A12;
hence ex B being set st
( B in S & ex C being thin of M st A = B \/ C ) by A12; :: thesis: verum

end;

then reconsider D = D as non empty Subset-Family of X by A1, A2, A10;
take D ; :: thesis: 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 ) )
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 A3; :: thesis: verum