let X, Y be set ; :: thesis: ( (proj1 X) \ (proj1 Y) c= proj1 (X \ Y) & (proj2 X) \ (proj2 Y) c= proj2 (X \ Y) )
thus (proj1 X) \ (proj1 Y) c= proj1 (X \ Y) :: thesis: (proj2 X) \ (proj2 Y) c= proj2 (X \ Y)

proof

let x be set ; :: according to TARSKI:def 3 :: thesis: ( not x in (proj1 X) \ (proj1 Y) or x in proj1 (X \ Y) )
assume A1: x in (proj1 X) \ (proj1 Y) ; :: thesis: x in proj1 (X \ Y)
then consider y being set such that
A2: [x,y] in X by RELAT_1:def 4;
not x in proj1 Y by A1, XBOOLE_0:def 5;
then not [x,y] in Y by RELAT_1:def 4;
then [x,y] in X \ Y by A2, XBOOLE_0:def 5;
hence x in proj1 (X \ Y) by RELAT_1:def 4; :: thesis: verum

end;

let y be set ; :: according to TARSKI:def 3 :: thesis: ( not y in (proj2 X) \ (proj2 Y) or y in proj2 (X \ Y) )
assume A3: y in (proj2 X) \ (proj2 Y) ; :: thesis: y in proj2 (X \ Y)
then consider x being set such that
A4: [x,y] in X by RELAT_1:def 5;
not y in proj2 Y by A3, XBOOLE_0:def 5;
then not [x,y] in Y by RELAT_1:def 5;
then [x,y] in X \ Y by A4, XBOOLE_0:def 5;
hence y in proj2 (X \ Y) by RELAT_1:def 5; :: thesis: verum