for x being Element of F1() holds
( x in F2() iff x in F3() )
proof
let x be
Element of
F1();
( x in F2() iff x in F3() )
hereby ( x in F3() implies x in F2() )
assume
x in F2()
;
x in F3()then
P1[
x]
by A1;
hence
x in F3()
by A2;
verum
end;
assume
x in F3()
;
x in F2()
then
P1[
x]
by A2;
hence
x in F2()
by A1;
verum
end;
hence
F2() = F3()
by Th8; verum