A1: for O being Operation of X st O = O2 AND holds
for L being List of X holds L | O = union { ((x . O1) AND (x . O2)) where x is Element of X : x in L }
proof
let O be Operation of X; :: thesis: ( O = O2 AND implies for L being List of X holds L | O = union { ((x . O1) AND (x . O2)) where x is Element of X : x in L } )
assume A2: O = O1 /\ O2 ; :: thesis: for L being List of X holds L | O = union { ((x . O1) AND (x . O2)) where x is Element of X : x in L }
defpred S1[ set , set ] means ( [\$1,\$2] in O1 & [\$1,\$2] in O2 );
let L be List of X; :: thesis: L | O = union { ((x . O1) AND (x . O2)) where x is Element of X : x in L }
thus L | O c= union { ((x . O1) AND (x . O2)) where x is Element of X : x in L } :: according to XBOOLE_0:def 10 :: thesis: union { ((x . O1) AND (x . O2)) where x is Element of X : x in L } c= L | O
proof
let z be object ; :: according to TARSKI:def 3 :: thesis: ( not z in L | O or z in union { ((x . O1) AND (x . O2)) where x is Element of X : x in L } )
assume z in L | O ; :: thesis: z in union { ((x . O1) AND (x . O2)) where x is Element of X : x in L }
then consider y being object such that
A3: ( [y,z] in O & y in L ) by RELAT_1:def 13;
reconsider y = y, z = z as Element of X by ;
( [y,z] in O1 & [y,z] in O2 ) by ;
then ( z in y . O1 & z in y . O2 ) by RELAT_1:169;
then ( z in (y . O1) AND (y . O2) & (y . O1) AND (y . O2) in { ((x . O1) AND (x . O2)) where x is Element of X : x in L } ) by ;
hence z in union { ((x . O1) AND (x . O2)) where x is Element of X : x in L } by TARSKI:def 4; :: thesis: verum
end;
let z be object ; :: according to TARSKI:def 3 :: thesis: ( not z in union { ((x . O1) AND (x . O2)) where x is Element of X : x in L } or z in L | O )
assume z in union { ((x . O1) AND (x . O2)) where x is Element of X : x in L } ; :: thesis: z in L | O
then consider Y being set such that
A4: ( z in Y & Y in { ((x . O1) AND (x . O2)) where x is Element of X : x in L } ) by TARSKI:def 4;
consider x being Element of X such that
A5: ( Y = (x . O1) AND (x . O2) & x in L ) by A4;
A6: ( z in x . O1 & z in x . O2 ) by ;
reconsider z = z as Element of X by A4, A5;
( [x,z] in O1 & [x,z] in O2 ) by ;
then [x,z] in O by ;
hence z in L | O by ; :: thesis: verum
end;
let O be Operation of X; :: thesis: ( O = O2 AND iff for L being List of X holds L | O = union { ((x . O1) AND (x . O2)) where x is Element of X : x in L } )
thus ( O = O2 AND implies for L being List of X holds L | O = union { ((x . O1) AND (x . O2)) where x is Element of X : x in L } ) by A1; :: thesis: ( ( for L being List of X holds L | O = union { ((x . O1) AND (x . O2)) where x is Element of X : x in L } ) implies O = O2 AND )
assume A7: for L being List of X holds L | O = union { ((x . O1) AND (x . O2)) where x is Element of X : x in L } ; :: thesis: O = O2 AND
now :: thesis: for L being List of X holds L | O = L | (O1 /\ O2)
let L be List of X; :: thesis: L | O = L | (O1 /\ O2)
thus L | O = union { ((x . O1) AND (x . O2)) where x is Element of X : x in L } by A7
.= L | (O1 /\ O2) by A1 ; :: thesis: verum
end;
hence O = O2 AND by Th30; :: thesis: verum