Let us consider the following truth table

$$\begin{array}{| c|c | c |} \hline p & q & p \implies q \\ \hline T & T & T \\ \hline T & F & F \\ \hline F & T & T \\ \hline F & F & T \\ \hline \end{array}$$

I want to understand sufficiency condition and necessary condition from the above truth table;

From $p \implies q$, in general we say

1) $p$ is sufficient for $q$

2) $q$ is necessary for $p$

How to interpret the necessary and sufficient condition from the truth table?

My interpretation along with doubts:

The cropped truth table that always results in true is as follows

$$\begin{array}{| c|c | c |} \hline p & q & p \implies q \\ \hline T & T & T \\ \hline F & T & T \\ \hline F & F & T \\ \hline \end{array}$$

From the truth table, in order for $q$ to become true, it is sufficient if $p$ becomes true(row 1);

In order for $p$ to become true, $q$ has to be true (row 1);

Is this interpretation valid? If yes, What is the need to remove row 2 from the actual truth table?


The interpretation is valid.

Row two is removed from consideration, because that is when $p\to q$ is false while we are asking: what happens whenever it is true that $p\to q$?

When $p\to q$ is true, then $p$ is only true when $q$ is true, although $p$ may be false when $q$ is true.   So $q$ is necessary for $p$, but not sufficient.

When $p\to q$ is true, then $q$ is always true whenever $p$ is true, although $q$ may be true if $p$ is false. So $p$ is sufficient for $q$, but not necessary.

When $p\to q$ is false, we do not care what happens.

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Your analysis will work ... if you are a little more careful. In fact, notice how you justify both cases by merely referring to (row 1) ... that's a little fishy, since that row is symmetrical with regard to $p$ and $q$ and yet you draw an asymmetrical conclusion from it! So that suggests that you need to do some more work here.

Indeed, for $q$ being a necessary condition of $p$, you really need to point out that the only rows where $p$ is true are ones where $q$ is true, i.e. that there are no rows where $p$ is true and $q$ is false ... and so you really have to look at all rows. And note that this also shows the relevance of taking out exactly that row where $p$ is true and $q$ is false, for again, if that row was still in there, then $q$ would not be necessary for $p$

Likewise, for checking whether $p$ is a sufficient condition for $q$, you cannot just point to row 1. Again, you need to check other rows; if there was a row where $p$ is true and $q$ is false, then clearly the truth of $p$ would not be sufficient for the truth of $q$. And so again it is crucial that that very row was in fact removed in your cropped table.

Indeed, rather than looking at all the rows that are left in your cropped table, it may be better to argue for the sufficiency of $p$ for $q$, and the necessity of $q$ for $p$, on the very basis of what rows are not in the cropped table!

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  • $\begingroup$ So, whenever we are examining statements like $p$ is sufficient for $q$, we are basically concerned about the truth values of $p$ and $q$ instead of being concerned about the whole $ p \implies q$ implication? $\endgroup$ – kauray Mar 19 '18 at 15:59
  • $\begingroup$ @KaustabhaRay Well, you can do either way: If you find that the implication is true, then you can say that $p$ is sufficient for $q$. But if you just know the values of $p$ and $q$, you can also figure it out (of course, knowing the values of $p$ and $q$ allows us to figure out thew value of the implication ... but the point is that you don't need that step) $\endgroup$ – Bram28 Mar 19 '18 at 17:25
  • $\begingroup$ If we just looked at the value of the rows, in the second row, value of p is True and value of q is False. This is where I get confused, isn't p being True sufficient for q to be True as well if we don't use the value of the implication as a whole? $\endgroup$ – kauray Mar 20 '18 at 3:16
  • $\begingroup$ Did you mean by "but the point is that you don't need that step" that just using the values of $p$ and $q$ we can say whether $p$ is sufficient for $q$ or not since we $can$ find the values of the implication using $p$ and $q$ but we $do not$ have to find the value of the implication? @Bram28 $\endgroup$ – kauray Mar 23 '18 at 16:42
  • $\begingroup$ @KaustabhaRay Yes, that's exactly what I meant! $\endgroup$ – Bram28 Mar 23 '18 at 17:34

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