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

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 $P \Longrightarrow Q \equiv \color{green}{\lnot P \vee Q}$ ?

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"You do your homework or you'll fail the course" – Hagen von Eitzen Aug 25 '13 at 13:07
Re your Edit: don't do that. – Did Aug 25 '13 at 15:36
Because people answer your question as it is hence, if you change the question later on, then their answer appears off-topic. Isn't this obvious? – Did Aug 27 '13 at 16:07
@Did: Thanks for your answer which I shall heed. I had thought to relocate "Source 1" because I realised it didn't feel like a question or intuition. – LePressentiment Aug 29 '13 at 7:39
$((\lnot P)\lor Q))\iff (Q\implies P)$ – user254665 Jan 1 at 22:27

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"?

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).

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+! Nice example! However it suffers from the fact that the converse "if one dies then one stops eating" is also blatantly obvious, though not a logical consequence of the statement. – Marc van Leeuwen Aug 25 '13 at 13:28
@MarcvanLeeuwen True. Replaced by somewhat more gory example. – Did Aug 25 '13 at 15:31
@LePressentiment Please do not edit posts for petty personal typographical concerns without at least asking the author's advice. – Did Oct 3 '13 at 12:43
@Did Please inform me if you mind my edit. I changed only two things: 1. I copied and pasted my improved version of my question in my OP. 2. Per my OP, I changed 'fail the course' to the simpler 'fail'. – LePressentiment Jan 1 at 21:41
@LePressentiment I do mind (unsurprisingly). – Did Jan 1 at 22:58

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: $$P \rightarrow Q \iff \lnot(P \land \lnot Q) \equiv \lnot P \lor Q$$

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Needs a TU, nice TT +1 – Amzoti Aug 25 '13 at 14:15
You're very welcome! – amWhy Sep 20 '13 at 14:42
I do thank you for your answer, but it seems to correspond to a question that requests a formal proof, like math.stackexchange.com/a/243980/53259, rather than mine? – LePressentiment Jan 2 at 1:13
@LePressentiment: this is the intuition for why they are equivalent. – Carl Mummert Jan 2 at 19:35
@CarlMummert But it uses a Truth Table and formal Rules of Replacement; my OP asked for a verbal example. – LePressentiment Jan 3 at 3:10

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.
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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.

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Merci beaucoup. J'ai plussoyé. – LePressentiment Sep 20 '13 at 13:45

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

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 $Q \lor \neg P$.

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.

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Sorry, but would you please clarify how this answers my question? You are proving the equivalence, but I ask for intuition? – LePressentiment May 3 '15 at 19:50
I might have done so the day after I pays it, but I answered this almost two years ago. – Tac-Tics May 3 '15 at 20:14
I do thank you for your answer, but it seems to correspond to a question that requests a formal proof, like math.stackexchange.com/a/243980/53259, rather than mine? – LePressentiment Jan 2 at 1:13

This answer is predicated on user 'Did' 's, but is not a duplicate.

Why cannot the bolded be symbolised as:
$\color{darkred}{P \vee \lnot Q}$ = "You neglect your homework or you won't fail"?

If you neglect your homework, then you fail  (ie: $\color{green}{\lnot P \vee Q}$ )
leaves open the possibility that somebody does homework AND fails ( ie $P \wedge Q$ ).
In fact, $\color{green}{\lnot P \vee Q}$ says nothing about what happens when one does one's homework.

$\color{#C41E3A}{\text{You neglect your homework or you do not fail ($P∨¬Q$)}}$
states that it is impossible that anybody does not neglect their homework AND fails ( $\color{olive}{ \lnot[\lnot P \wedge Q]}$ ); but notice that by DeMorgan's Law, $\color{olive}{ \lnot[\lnot P \wedge Q]} \equiv \color{#C41E3A}{P∨¬Q}$.

As another intuitive example, consider this true statement: If one is decapitated, then one dies.
Here, P = "one is decapitated" and Q = "one dies";
$\color{green}{\lnot P \vee Q}$ = One is not decapitated or one dies.

For this Substitution Instance, $\color{#C41E3A}{P∨¬Q = \text{One is decapitated or one lives.}}$
But this is clearly wrong since most people die with their head still attached to their body.

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Hmm, here's my way of thinking about it. There are four things that can happen.

A) You can neglect your homework and you can fail.

B) You can neglect your homework and pass.

C) You can do your homework and fail.

D) You can do your homework and pass.

Let's imagine there are several teachers all trying to impart different messages.

\Large{1.}$Suppose you have a STRICT teacher, who says: "If you neglect your homework, then you WILL fail:$P \implies Q$." She is saying that B will never happen. A might happen, and indeed, if you neglect your homework then A is the ONLY possibility. B and C might happen.$\Large{2.}$Suppose you have an EXISTENTIAL teacher, who says: "Well, one of two things (or maybe both) are going to happen; either you are going to remember your homework or you are going to fail the class. That's just how it is:$-P \vee Q$." In this case: B ($P \wedge -Q$) is not possible. All the others are possible, as A, C, and D all have either$-P$(B and C have -P) or they have Q (A and C have Q), but only B lacks both$-P$and$Q$. B will never happen. A, C, and D might. But notice STRICT teacher and EXISTENTIAL teacher have the exact same outcome.$P \implies Q \quad \iff -P \vee Q\Large{3.}$Now we have NICE TEACHER, who says: If you do your homework you will pass; that's all I ask: -P => -Q. If you don't do your homework, then maybe you will pass and maybe you won't. The only way to fail is if you don't do your homework. If you fail, that means you didn't do your homework:$Q => P$In this case C is the only option that is impossible. All others are possible. (If you do your homework, it is impossible to fail.) -P => Q <=> Q => P.$\Large{4.}$Then there is OBSCURE teacher, who says: P V -Q. One of two things will happen. Either you will forget your homework (P) or you will pass the class (-Q)". So either P happens (A or B) or -Q happens (B, or D). The only thing that can not happen is C in which neither P nor -Q occurs. So OBSCURE teacher is saying the same thing as NICE teacher:$P \vee -Q \quad \iff \quad -P \implies -Q \quad \iff \quad Q \implies P\$.

There's also PUSHOVER teacher "You will pass the course no matter what: -Q" SADIST teacher "You will fail the course no matter what: Q". DIDACTIC teacher "You will pass this course if and only if you do your homework: (P V Q) OR (-P V -Q)" and INSANE teacher "You will pass this course only if you neglect your homework" and so on.

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Then there is OMNIPOTENT teacher who say's "I am going to make it so that all of you will do your homework and all of you will pass: -P V -Q. Only A is possible". – fleablood Jan 1 at 23:49
+1. I will reread this for better comprehension, but please inform me if you mind my changes (to improve readability). Thanks. – LePressentiment Jan 2 at 1:09