Q:

Logic coursework assistance

I'm doing some homework for my summer logic course, and I ran into something of a roadblock. I dont want to post the actual homework problem, so I found two problems from past exercises that, if someone were to explain how these were done, would probably help me solve my homework. See: http://imgur.com/KC5FEOv
My study group is completely stumped. We think it has something to do with creating a truth table, but there must be another step, as the TA informs us that we're 'missing something'. Can anyone provide a nudge in the right direction?
edit: a similar but different problem is available on the course website. It provides an answer, but shows no work. Explaining what's going on here would be similarly helpful. see: http://imgur.com/sEwcEWV

A:

I'll restate the first problem in slightly different notation and then give a solution.
Problem #1: Let $D=\{a,b\}$ be the domain for two parameters $x$ and $y$. [I will assume that $a$ and $b$ are distinct. I believe the author of the problem intends this.] We are to show that $$ \exists x,y \in D, P(x,y) \wedge Q(x,y) \not= \left( \exists x,y\in D, P(x,y) \right) \wedge \left(\exists x,y\in D, Q(x,y) \right)$$
for some predicates $P$ and $Q$. In other words, we are trying to show that you can't just distribute existential quantifiers across conjunctions.
It makes sense that these are not equal. The LHS (left hand side) forces us to use the same $x$ in both $P$ and $Q$ (and the same $y$ in both $P$ and $Q$) while the RHS (right hand side) allows different choices for each predicate. 
Really the goal is to cook up some $P$ and $Q$ that cause the above statement's LHS to mismatch the RHS. The easiest ones that I could think of are: $P(x,y)$ is "$x=a$" and $Q(x,y)$ is "$x=b$". 
In this case, the LHS is false. Why? If there exists $x,y \in D$ such that $P(x,y)\wedge Q(x,y)$ is true, then both $P(x,y)$ and $Q(x,y)$ are true. So we have both $x=a$ and $x=b$ so that $a=b$ (which goes against the assumption that $a$ and $b$ are distinct).
On the other hand, the RHS is true. Why? "$\exists x,y\in D, P(x,y)$" is seen to be true by letting $x=a$ and $y$ be whatever AND "$\exists x,y\in D, Q(x,y)$" is seen to be true by letting $x=b$ and $y$ be whatever. So both are true thus their conjunction is true.
By the way, another solution would be: Let $P(x,y)$ be "$x=y$" and $Q(x,y)$ be "$x \not= y$". Do you see why this works as well?
I hope this helps you get past your roadblock!

