# 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?

-

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:

On a side note, for future reference, using $\LaTeX$ would probably be a better alternative to inserting mathematical notation in your posts on SE.