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?
Best Answer
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:
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.