Express "Someone has visited every country except Libya" using predicates and quantifiers. I have expressed the statement "Someone has visited every country except Libya" using quantifiers as follows.
$$\exists x\forall y [V(x, y)\wedge \sim V(x, Libya))]$$
where $V(x, y)$ represents $x$ has visited country $y$.
Is the above representation correct?
Note: Domain for $x$ is all people in this world, and for $y$ is all countries.
 A: No, it is not exactly correct, as pointed out by Michael in the comments. I'd write it as $$\exists x\forall y,\;\lnot V(x,y)\leftrightarrow y=Libya$$
A: 
Is the above representation correct?

No. This representation is not exactly correct. To understand why, first see the following statement taken from here.

A sentence ∀xP(x) is true if and only if P(x) is true no matter what value (from the universe of discourse) is substituted for x.

According to this statement, your representation ∃x∀y[V(x,y)∧∼V(x,Libya))] is true if and only if "V(x,y)∧∼V(x,Libya))" is true for all values of y. But, if we take y = Libya, this expression becomes "V(x,Libya)∧∼V(x,Libya))" which is false making your representation false too.
To prevent your expression from becoming false, you need either ∨(disjunction) or →(implication). Based on these two options, you can represent this statement as follows:

*

*By using Implication: 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".

*By using Disjunction: you can 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))]. You can simplify it as ∃x∀y[(B(x,y)∧(y≠Libya))∨(¬B(x,Libya))] which is very close to your representation.

All these expressions are equivalent and inter-convertible.
