Correct Form of a Logical Statement I ran across a problem which has stumped me involving existential quantifiers.
Let U, our universe, be the set of all people. Let S(x) be the predicate "x is a student" and I(x) be the predicate "x is intelligent".
I want to write the statement "Some students are intelligent" in the correct logical form. I can see 2 possible ways to write it
1)  There exists an x in U such that ( S(x) AND I(x) )
2)  There exists an x in U such that ( S(x) implies I(x) )
If I draw a Venn diagram, it seems like option 1 must be true, but from this same diagram (where the sets where S(x) is true and I(x) is true intersect), it is also true that there is an x such that if x is in the set where S(x) us true, then x is in the set where I(x) is true. This makes me wonder if these two statements are not logically equivalent, but I have a feeling they are not.
Thanks,
Matt
 A: The two statements are not equivalent. The second one would be true if there is even one nonstudent $x$ in the universe, regardless of intelligence; it would also be true if there is even one intelligent person $x$ in the universe, regardless of student status.
Understanding why this is the case depends on truly understanding how mathematical if-then statements work (in this case, the existential quantifier can be ignored, as I don't think it's part of the error you're making).
A: The confusion lies in choice of logical connective to represent a restriction  on the domain of discussion.   We use conjunction to restrict an existential quantifier, and implication to restrict a universal quantifier.
Here we are restricting the domain, from discussions of all people in the given universe, to students in that universe.



*

*$\exists x \in U \big(S(x)\wedge I(x)\big)$


"There exists $x$ in $U$ such that $S(x)$ and $I(x)$" is "someone in our universe is both a student and intelligent" ie "some students are intelligent".
We use conjunction as the connective for the restricted existential because to be true there merely needs be an example of a person who is a student and intelligent.
If our universe consists of teachers Bob and Jane, and students Tom, Dick, and Hariet, then our statement "some students are intelligent" will only be true if at least one of Tom, Dick, or Harriet is intelligent.



*

*$\forall x\in U\big(S(x)\to I(x)\big)$


"For all $x$ in $U$ it is such that $S(x)$ implies $I(x)$" is "everyone in our universe, is intelligent whenever they are a student"  ie: "every student is intelligent."
We use implication as the connective for the restricted universal because to be true then every person must either be intelligent or be not a student.   That is, $\;\forall x\in U\big(\neg S(x)\vee I(x)\big)\;$.
If our universe consists of teachers Bob and Jane, and students Tom, Dick, and Harriet, then our statement "all students are intelligent" will only be true if all of of Tom, Dick, and Harriet are intelligent; that is if being a student implies being intelligent.

We further note that $\;\exists x\in U\big(S(x)\to I(x)\big)\;$ can be true only if there is someone in that universe who is either intelligent or is not a student.
Likewise $\;\forall x\in U\big(S(x)\wedge I(x)\big)\;$ can be true only if everyone in that universe is both a student and intelligent.
Don't use the wrong connective to represent a restrictive quantification.
A: The statement "some students are intelligent" can be rephrased as "there is at least one student who is intelligent".  So your logical statement 1) is correct.  However, 2) is not correct and 2) is not logically equivalent to 1).  
Recall that "p implies q" is false exactly when p is true but q is still false.  But if p is false, then regardless of the truth value of q, "p implies q" is true.  Thus, if S(x) is false, then "S(x) implies I(x)" is true. Thus, if there is an x in U such that S(x) is false, then "there is an x in U such that (S(x) implies I(x))" becomes true.
