# Intuition behind “If P then Q” = “Q or Not P ”

I understand with truth tables the Conditional Law: $[P \Longrightarrow Q] \equiv [\lnot P \vee Q]$. However, what's the intuition or a natural motivation? Source 1, all but intuitive, now appears as an Answer beneath. Retrospectively, cogent and instructive were Cameron Buie's and Prof Scott's Answer.

Source 2: P46 of How to Prove It by Daniel Velleman :

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

But what message is the teacher trying to convey with this statement?
Clearly the intended message is “If you neglect your homework, then you’ll fail the course,” 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.

Based upon “If you neglect your homework, then you’ll fail the course,”
could I not aver its "intended meaning" as
$P \vee \lnot Q$ = "You neglect your homework or you won't fail the course"?
This would fail to motivate the Conditional Law.

Source 3 : Ex 2.16 on P48 of Gary Chartrand et al's Mathematical Proofs, 2nd Ed:

"If you earn an A on the final exam, then you will receive an A for the.final grade."

We need know only that the student did not receive an A on the final exam or the student received an A as a final grade to see that she kept her promise.

I can't twig this due to its terseness. Could the details please be explicated?

-
"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

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).

-
+! 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: I beg for your forgiveness. Would you mind my edit? –  LePressentiment Oct 27 '13 at 13:49

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

-
Needs a TU, nice TT +1 –  Amzoti Aug 25 '13 at 14:15
Thank you very much. I'm not admitted of multiple best answers, so I upvoted. –  LePressentiment Sep 20 '13 at 14:39
You're very welcome! –  amWhy Sep 20 '13 at 14:42

Source 1 : http://plus.maths.org/content/os/issue36/features/nishiyama/index.
I understand this argument, but it does not feel intuitive.

We are given that $\color{green}{P \Longrightarrow Q}$.
$\boxed{\text{Case 1 of 2 : } \color{green}{P \Longrightarrow Q} \text{ true}}$ Substitute the green into the tautology $\color{green}{P} \vee \lnot P$ to get: $\color{green}{Q} \vee \lnot P$. $\boxed{\text{Case 2 of 2 : } \color{green}{P \Longrightarrow Q} \text{ false}}$ Since an argument is valid if the premises cannot all be true without the conclusion being true as well, thus this case means $[\text{$P$is true$\& \, Q$is false]} \iff [\text{$\lnot P$is false$\& \, Q$is false]} \iff [\text{$\lnot P \vee Q$is false]}$.

Altogether, the 2 cases result in: $\left\{ \begin{array}{rcl} \color{green}{P \Longrightarrow Q} \text{ true$\equiv [\lnot$P$\vee$Q] is true} \\ \color{green}{P \Longrightarrow Q} \text{ false$\equiv [\lnot$P$\vee$Q] is false} \end{array}\right.$.
Thus, the Conditional Law has been proven.

-
What is going on? –  Did Aug 25 '13 at 15:34

This answer is predicated on Did's, but is not a duplicate.
To obviate litotes, I define "accomplish = not neglect".

Based upon “If you neglect your homework, then you’ll fail the course,” could I not assert that its "intended meaning" is $\color{#C41E3A}{\text{$P∨¬Q$= "You neglect your homework or you won't fail the course"}}$?

If you neglect your homework, then you’ll fail the course ( ie $\lnot P \vee Q$ ) leaves open the possibility that somebody accomplishes homework AND fails the course ( ie $\lnot P \wedge Q$ ).
In fact, $\lnot P \vee Q$ says nothing about what happens when one accomplishes one's homework. $\color{#C41E3A}{\text{You neglect your homework or you won't fail the course ( ie$P∨¬Q$)}}$ states that
it's impossible for someone to accomplish his/her homework AND fail the course.

Consider the statement that If one is decapitated, then one dies. You probably agree it is true.
Here, P = "one is decapitated" and Q = "one dies".
For this instantiation, $\color{#C41E3A}{P∨¬Q = \text{One is decapitated or one lives.}}$
But this is clearly wrong since there exists people who die with their head still attached to their body (most of them, actually).

-

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.

-
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.

-
Thank you. I upvoted. –  LePressentiment Sep 20 '13 at 13:41

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.
-