My textbook states that for the conditional statement "p implies q",

"p is a sufficient condition for q and q is a necessary condition for p."

How is this so? One might be lead to believe that p is independent of q and that it is a necessary and sufficient condition for q, no?

Additionally, what is the correct way to interpret the truth table of this conditional statement given below:

TT for conditional statement

Edit: Please keep in mind that I have only just graduated high school and am not taking any advanced courses in logic.

  • $\begingroup$ (it's generally better to ask one question per post; if you have two questions, make two posts!) $\endgroup$ – user14972 Sep 6 '16 at 1:46
  • $\begingroup$ Oh, OK. I just thought that it will be easy for the answerers to slip in some information regarding the truth table as well. Will keep this in mind for future posts. $\endgroup$ – user361896 Sep 6 '16 at 1:47
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    $\begingroup$ @KaumudiHarikumar The interpretation is that the conditional statement is only contradicted when $p$ is true but $q$ is false. $\endgroup$ – Graham Kemp Sep 6 '16 at 2:36
  • $\begingroup$ @GrahamKemp: I have understood this now. Thanks :) $\endgroup$ – user361896 Sep 6 '16 at 2:39
  • $\begingroup$ You may find my answer to a similar question at math.stackexchange.com/questions/1551320/… to be helpful in understanding logical implications. $\endgroup$ – Dan Christensen Sep 6 '16 at 2:57

It's important to realize that $p \implies q$ is not any kind of causative relationship - it's not saying $p$ causes $q$, or anything like that. It's saying "if $p$ happens to be true, then $q$ also happens to be true, possibly by pure coincidence".

Likewise, "necessary condition" and "sufficient condition" don't say anything about causation. $q$ is a necessary condition for $p$ exactly if $p$ can only be true if $q$ is; $p$ is a sufficient condition for $q$ exactly if whenever $p$ is true, so is $q$ - possibly coincidentally.

So suppose we know that the sentence $p \implies q$ is true. Then we know that if $p$ happens to be true, so does $q$; that's what the $\implies$ symbol means. So $p$ can't be true if $q$ isn't - if it were the case that $p$ is true with $q$ false, our sentence $p \implies q$ would be false. But that's the definition of $q$ being a necessary condition for $p$! Likewise, $q$ has to be true whenever $p$ is, which means $p$ is a sufficient condition for $q$.

Notice that it doesn't work the other way around - $p \implies q$ doesn't mean $p$ is a necessary condition for $q$, because $q$ might be implied by other, unrelated things as well.

We remind that if $p \implies q$, then: - $q \implies - p$


These are other equivalent ways to say $p \implies q$ in English. Personally I think "$p$ implies $q$" is the clearest, but I suppose it is important to understand the other phrases, since people use them often. Here is the intended interpretation.

"$p$ is a sufficient condition for $q$": if you want to know if $q$ is true, then it is enough (sufficient) to know that $p$ is true. However, there may be other ways for $q$ to be true, so in this case you cannot say $p$ is a necessary condition for $q$. [As an example, if $p$ is "my favorite number is divisible by $4$" and if $q$ is "my favorite number is even," then knowing $p$ is true is enough (sufficient) to conclude that $q$ is true, but there are other ways for $q$ to be true without $p$ being true, e.g. the case where my favorite number is $2$.]

"$q$ is a necessary condition for $p$": simply, if $p$ is true, then $q$ must (necessarily) also be true. It is necessary in the sense that if $q$ were not true, then it would be impossible for $p$ to be true.


Sufficient: It is enough for $p$ to be true in order to conclude that $q$ is also true.

Necessary: If $q$ is not true, then $p$ has to be false. For if $p$ were true, it would imply that $q$ is also true, a contradiction.

Regarding your second question, the idea is that, when the antecedent is false, the conditional is "vacuously" true. For instance, the pythagorean theorem says that if a triangle has a right angle, then the square of the hypothenuse equals the sum of the squares of the other two sides. If we're not dealing with a right triangle, then its meaningless to apply this theorem, so we cannot claim it is false. To give an extreme example, the sentence:

"If D is a duck, then the square of its hypothenuse equals the sum of the squares of the other two sides."

Seems rather irrelevant to the validity of this theorem, and claiming it is false because that sentence is not true, isn't too reasonable.

  • $\begingroup$ Okay, but I don't understand this completely. Just as it is possible to reason that if the antecedent is false, it doesn't make sense to say that the consequent will be false, how does it make sense to assert that it is "vacuously" true? (Forgive me for my lack of knowledge; like I mentioned in the question, I am not taking any advanced courses in logic and graduated high school just this year.) $\endgroup$ – user361896 Sep 6 '16 at 1:53

If you think of $p \implies q$ being equivalent to the statement $\lnot p \lor q$ (verify this with your truth table), that might help clarify what that wordy statement is trying to say.

It is essentially saying that there are two ways in which the statement $p \implies q$ is true. Either when $p$ is not true or when $p$ is true and $q$ is true.

p is a sufficient condition for q: This means that if $P$ is true, $Q$ is guaranteed to be true. In other words, it is sufficient for $P$ to be true to guarantee that $Q$ is true.

q is a necessary condition for p: Essentially means $P$ can't be true without $Q$ being true. It is necessary that $Q$ is true for $P$ to be true but not sufficient because there can be a case where $Q$ is true but $P$ is false. In other words, if $Q$ is false, then $P$ must be false. but if $Q$ is true, you cannot guarantee that $P$ is true (this follows from the contrapositive: $\lnot P \implies \lnot Q$).

Edit: Why is the statement true if the antecedent is false?

When you make a logical statement such as $P \implies Q$, you are saying this entire statement is something that is always true. Thus under any circumstance, the statement must be true. This is where a lot of people get confused so an example would clarify.

If I make the statement for all $x$, $ \text{"if x is a puppy"} \implies \text{"x is cute"}$, then I am saying this statement is a correct statement in that it must always be true. Now let us consider different cases:

  1. x is a puppy - Well we claim if x is a puppy, it must be cute and in this case x is a puppy so it is necessary that it is cute. If x is a puppy and it isn't cute, then we have messed up because the logical statement we made must always be valid! This is important to realize; there is no such situation where the logical claim you made should should be false and in this case, the claim you made is that if x is a puppy, it must be cute.
  2. x is not a puppy - Now even in this case, our logical statement must stand. This is the case you are interested in because you asked, why is the statement true if the antecedent is false. Like I said, the statement must always be true and so whoever made $P \implies Q$, defined it so that it will be true in the case that P is not true. Think of it this way; if x is not a puppy, then we can make no conclusions about whether or not x is cute. x could be a vulture in which case it is not cute, or x could be a cat in which case it is cute. Either way, our logical statement needs to be valid and thus it is just defined as being true when x is not a puppy. It is not so much that the statement is true when the antecedent is false as it is that the statement is defined to be true when the antecedent is false.

In summary, when you make the statement $P \implies Q$, you are saying this statement is universally true. The claim that the statement is making is that whenever $P$ is true, you can safely conclude $Q$ is true and whenever $P$ is not true, you can't say anything about $Q$, and this is an airtight statement that is never going to be invalid or false for any case. So if somebody tries to disprove you by saying "hey, I found a case where P is false but Q is true, your statement is wrong", you can simply say that your statement makes no claims that if $P$ is false then $Q$ must be as well and thus even in this case, is a valid statement.

  • $\begingroup$ I don't understand how the statement can be true even when the antecedent is false. (I know that there are some other posts dealing with the same question but since I am not well educated in the field of logic, I find those posts a little difficult to understand.) $\endgroup$ – user361896 Sep 6 '16 at 2:00
  • $\begingroup$ @KaumudiHarikumar No worries; ask for any clarification. I updated my answer to try and detail why the statement is true when the antecedent is false. $\endgroup$ – gowrath Sep 6 '16 at 2:27
  • $\begingroup$ Yes, I am reading your edited answer now and I have one doubt; isn't $p→q$ equivalent to ~$p^q$? (~$p$ AND q?) $\endgroup$ – user361896 Sep 6 '16 at 2:44
  • $\begingroup$ @KaumudiHarikumar Try make a truth table for $\lnot p \land q$ and see if it matches the truth table in your question. Not P AND Q means the statement is true only if P is false and Q is true. That does't sound right does it? $\endgroup$ – gowrath Sep 6 '16 at 2:47
  • $\begingroup$ Oh, sorry! That was supposed to be $p$^~$q$! My bad. $\endgroup$ – user361896 Sep 6 '16 at 2:53

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