Can $P \implies Q$ be represented by $P \vee \lnot Q $?

Question

_{Source: p 46, How to Prove It by Daniel Velleman}

Though the author writes $Q$ (the original apodosis) as 'You'll fail the course',
for brevity I shorten $Q$ to 'You fail'.

Let $P$ be the statement “You will neglect your homework” and $Q$ be “You fail.”
Then “You won’t neglect your homework, or you fail.” $ \quad = \color{green}{\quad \lnot P \vee Q}$.

But what message is the teacher trying to convey with this statement?
Clearly the intended message is
“If you neglect your homework, then you fail,” or in other words $P \rightarrow Q.$
Thus, in this example, the statements $\lnot P \vee Q$ and $ P \rightarrow Q $ seem to mean the same thing.

Why cannot the bolded be symbolised as
$\color{darkred}{P \vee \lnot Q}$ = "You neglect your homework, or you fail not." ?
I am trying to intuit Material Implication: intuitively, how does $P \Longrightarrow Q \equiv \color{green}{\lnot P \vee Q}$ ?

Answer 1

The part of your post where I could find a real question is:

Based upon “If you neglect your homework, then you’ll fail the course,” could I not assert that its "intended meaning" is P∨¬Q = "You neglect your homework or you won't fail the course"?

Answer:

No you could not. If you neglect your homework, then you’ll fail the course leaves open the possibility that somebody does not neglect their homework AND fails the course (in fact this says nothing about what happens when one does not neglect one's homework). You neglect your homework or you won't fail the course states that it is impossible that anybody does not neglect their homework AND fails the course.

Consider the statement that If one's head is severed, then one dies. You probably agree it is true. Now the analogue of your suggestion is: One is decapitated or one lives. But this is clearly wrong since there exists people who die with their head still attached to their body (most of them, actually).

Answer 2

Perhaps one way to look at the equivalence of $P \rightarrow Q$ is to note that by definition, (and confirmed with a truth table, below, compliments of Wolfram Alpha),

$$P \rightarrow Q\;\text { is TRUE $\;$ if and only if}\; \;(P \;\text{is false $\;$ or $\;$} \;Q \;\text{is true})$$

Translating, $$\underbrace{P \rightarrow Q}_{\text{is true}} \iff (\underbrace{\lnot P}_{P\; \text{is false}} \lor \underbrace{Q}_{Q\;\text{is true}})$$

For this reason, we can also say that $P \rightarrow Q$ is true if and only if it is not the case that $P$ is true and $Q$ is false: $$P \rightarrow Q \iff \lnot(P \land \lnot Q) \equiv \lnot P \lor Q$$

Answer 3

The most intuitive view of this for me in one sentence is: $\color{ #FF4F00}{Q} \wedge \lnot P$.

1. Either $\color{ #FF4F00}{Q}$ is true.
2. Or $Q$ is not true. In this second case, then $P$ is not true. This is merely the contrapositive of the given postulate: $P$ implies $Q$.

Source 2 just instantiates the above general case :

1. Either $\color{ #FF4F00}{\text{you have failed the course.}}$
2. Or if you didn't fail, then you didn't neglect your homework. This is merely the contrapositive of the given postulate: Neglect of homework $\Longrightarrow$ Failure.

Answer 4

I think such "real world" examples generally only confuse your intuition, since there is usually more conveyed than just the logical content of the statement. For instance saying $P\Rightarrow Q$ in real world situation often suggests that $P$ is (possibly) realised before $Q$, but that is not part of the logical content. For instance this is the case for "if you neglect your homework then you will fail the course". Indeed at a point in time when $P$ is realised but $Q$ is not yet (I neglected my homework but did not yet fail the course) one is precisely in the situation the statement $P\Rightarrow Q$ claims cannot exist. I'll leave the question of whether or not the statement "you will fail the course" already has a truth value at that point to philosophers.

So it is better to use more neutral, timeless examples, such as "if $n$ is a prime number, then $n$ is positive". This (rather bland) statement means the same as "either $n$ is not a prime number or $n$ is positive", which is $\lnot P\lor Q$ here; note that there is room for both parts of the "or" to be true (which is fine, because it is not an exclusive or) but not for both parts to be false (that would require a negative prime number), which is what "or" means.

In this case your $P\lor\lnot Q$ would be "either $n$ is prime or $n$ is not positive", but that fails for any positive composite number, so it is not valid, while $P\Rightarrow Q$ is valid.

Note also that the logical meaning does not involve causality; it would be hard to argue that $7$ being positive is caused by $7$ being a prime number.

Answer 5

This follows simply from the Law of Excluded Middle: $P \lor \neg P$ for all propositions $P$.

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Let's assume $P \rightarrow Q$ and deduce $Q \lor \neg P$.

By the Law of Excluded Middle, we have either $P$ or we have $\neg P$. We do a case analysis over which one is true:

If we got a $P$, by our assumption, we can deduce $Q$.

If we got a $\neg P$, well... we stick with $\neg P$.

Having covered both cases, we know one or the other is true, so we have $Q \lor \neg P$.

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Let's go the other way now, to show the statements are logically equivalent. We'll assume $Q \lor \neg P$ and deduce $P \rightarrow Q$.

We assume we have a $P$, and our goal is to produce a $Q$, and this will prove $P \rightarrow Q$.

Do case analysis on our assumption $Q \lor \neg P$.

If we have $Q$, we've have fulfilled our obligation.

If we have a $\neg P$, on the other hand, we notice that we also have a $P$ just hanging around in our current list of assumptions. This gives us a contradiction, and by Principle of Explosion, we may conclude $Q$. This again fulfills our obligation.