Making Sense of Implication vs. Conjunction

We've begun CSC165 with symbolic logic. I'm still working to get used to mathematical logic, which has exhibited differences to the logic we use on a day-to-day basis.

So far, one of the things that has confused me the most was what implication means. Particularly, why P(x) => Q(x) is equivalent to the expression (not P(x) and Q(x)), why it includes the area outside of P(x) and Q(x) and the rest of Q(x) that doesn't overlap with P(x), and why can't P(x) => Q(x) be expressed as (P(x) and Q(x)) since the two expressions mean the same thing.

P(x) => Q(x)

After seeking ample explanations, I finally come to terms with the use of implication vs conjunction:

- P(x) AND Q(x) implies everything in the universe is in P(x), and additionally includes elements that are ALSO Q(x) in addition to being P(x) with no additional associations (i.e. outside of these spaces) 

- P(x) IMPLIES Q(x) means you're acknowledging there are elements outside of P(x), that a subset P(x) exists in universe, and if an element is in the subset P(x) then it's also in Q(x), HENCE the area in Q(x) that doesn't overlap with P(x) is highlighted, and the area outside of both P(x) and Q(x) is also highlighted

I found a good example from Stack Overflow that further clarified this difference:

Q:

Some dogs bark ∃x (dog(X) Λ bark(x))
All dogs have four legs ∀x (dog(x) -> have_four_legs(x))
My question is: is it possible for the second example to be: ∀x (dog(x) Λ have_four_legs(x))
And why can't the first example be: ∃x (dog(X) -> bark(x))

A:

Well ask yourself this: Are implication and conjunction equivalent? No. Your last statement says that all x's are both dogs and have four legs. While that does mean that all dogs have four legs, it also means that everything is a dog.

http://stackoverflow.com/questions/5060247/when-to-use-conjunction-and-when-to-use-implication-first-order-logic