Deriving Universal Modus Tollens I'm asked to derive the validity of Universal Modus Tollens from the validity of Universal Instantiation and Modus Tollens.  I'm new to this deriving/proving stuff, so I'm not sure if I'm doing it right, but here's what I came up with:
Universal instantiation says that if (1) is true:
(1) ∀x, P(x) → Q(x)

Then (2) is true for any particular item y
(2) P(y) → Q(y)

Modus Tollens says that if (2) is true and (3) is true:
(3) ~Q(y)

Then (4) is true:
(4) ~P(y)

Therefore if (1) is true, and (3) is true, then (4) is true. In other words, the following argument is valid:
∀x, P(x) → Q(x)
~Q(y)
∴ ~P(y)

And that's Universal Modus Tollens.
Am I doing this right?  Am I making any unwarranted assumptions or unsupported claims?  Am I skipping any steps?
 A: Let's do a little logic magic, shall we?
Note: I will assume here "$\implies$" is identical with "$\rightarrow$", and therefore that you are asserting the proposition in (1) rather than merely considering it. See discussion here for a reason why.
Let the set $S$ be the set of those elements, $x$, such that statements such as "$P(x)$" or "$Q(x)$"  make sense and for such statements it makes sense to consider the propositions of the form "$P(x) \implies Q(x)$".
Consider then, since:
$$
\tag{1} \forall x \in S,P(x) \implies Q(x),
$$
$$
(1) \iff \tag{2} \forall x \in S, (\text{~}P(x) \vee Q(x)) \text{ is true.}
$$
Asserting:
$$
\tag{3} \exists y \in S, \text{~}Q(x) \text{ is true},
$$
we get,
$$
\tag{4} (\text{~}Q(x) \text{ is true}) \implies (\text{~}P(x) \text{ is true}).
$$
For the sake of brevity, I'm skipping a step here; though it is admittedly trivial.
Therefore,
$$
\tag{5} (1) \wedge (3) \implies \text{~}P(x) \text{ is true}.
$$
Q.E.D.
EDIT:
So, to answer your questions:


*

*You have this right.

*For most purposes, you have no problem with assumptions.

*Finally, you skipped a step between (3) and (4) in your post, which was made explicit by (2) in mine.


On a side note, for future reference, using $\LaTeX$ would probably be a better alternative to inserting mathematical notation in your posts on SE.
TIP: You can tag your equations by using "\tag{n}", where "n" denotes the number or string you want to tag your equation with.
A: I apologize for being late to the game, but here goes:

Since  due to the identity rule, then . Now distribute  to get .  
 is a contradiction, or a negation law. Thus .   So we now have .   we can now use the identity law of , assuming . Thus  
 is the definition of Decomposing a conjunction in the laws of Tautology. Please refer to page 7 of the following link: http://sites.millersville.edu/bikenaga/math-proof/truth-tables/truth-tables.pdf
