Writing "Someone has visited every country in the world except Libya" using quantifiers. Someone has visited every country in the world except Libya.
Let $B(x,y)-x\,has\,visited\,country\,y$
Domain for x is all people and for y is all countries
is $\exists x \forall y [B(x,y) \land (y \neq Libya)]$
the correct representation?
 A: Your original statement implies that the someone in question has not visited Libya. That is, we need $B(x$, Libya$)$ to be false, but your answer doesn't imply that. According to your answer, $B(x$, Libya$)$ may be true or false, since the expression will evaluate to false because of $y\ne$ Libya regardless. You can modify it as:
$\exists x[\forall y (B(x,y) \land (y \neq $Libya$))]\wedge \neg B(x$, Libya$)$
A: The expression ∃x∀y[B(x,y)∧(y≠Libya)] is incorrect because it implies "There exists a person for whom no country is Libya and who has visited every country in the world". As obvious, "no country is Libya" is not what you want to imply. It is a common mistake in case of universal quantifier(∀) to "universalize" the subseqent proposition.
You can use the conditional operator in this case in order to prevent the "universalization". What you can do is to represent the given statement as "There exists a person who, for all countries in the world, if a country is not Libya, then he has visited that country and if he has visited a country then that country must not be Libya". i.e. ∃x∀y[((y≠Libya)→B(x,y)) ∧ (B(x,y)→(y≠Libya))] which is same as ∃x∀y[(y≠Libya)↔B(x,y)] or "There exists a person who, for all the countries in the world, has visited a country if and only if that country is not Libya". This is same as the given statement "Someone has visited every country except Libya".
If you don't want to use the conditional operator, you can also represent it as "There exists a person who, for all countries in the world, has either visited the country or that country is Libya".i.e. ∃x∀y[(B(x,y)∧(y≠Libya))∨(¬B(x,y)∧(y=Libya))].
Both of these are equivalent and hence inter-convertible.
