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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)
∴ ~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?

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up vote 1 down vote accepted

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:

$$ \tag{1} \forall x \in S,P(x) \implies Q(x), $$

$$ (1) \iff \tag{2} \forall x \in S, (\text{~}P(x) \vee Q(x)) \text{ is true.} $$


$$ \tag{3} \exists y \in S, \text{~}Q(x) \text{ is true}, $$

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.


$$ \tag{5} (1) \wedge (3) \implies \text{~}P(x) \text{ is true}. $$



So, to answer your questions:

  1. You have this right.
  2. For most purposes, you have no problem with assumptions.
  3. 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.

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I edited the 3rd list item, I had referred to the wrong lines. – ThisIsNotAnId Apr 29 '12 at 17:37

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:

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