My question pertains to pages 44-46 of section 1.5 - The Conditional and Biconditional of How to Prove It by Velleman.
Construction of the truth table for $P \implies Q$ begins at the bottom of pg. 44. Velleman starts by pointing out that since $P \implies Q$ stands for the statement "if $P$ is true, then $Q$ is true", this statement is clearly false if $Q$ is false when $P$ is true. Similarly, this statement is clearly true in the case where both $P$ and $Q$ are true, so we have:
$$\begin{array}{|c|c|c|} \hline P&Q&P\Rightarrow Q\\ \hline F&F&?\\ F&T&?\\ T&F&F\\ T&T&T\\ \end{array}$$
This is clear to me, but I'm not sure I understand the gist of Velleman's explanation for lines 1 and 2 (emphasis added):
To help us fill in the undetermined lines in this truth table, let’s look at an example. Consider the statement “If $x > 2$ then $x^2 > 4$,” which we could represent with the formula $P(x) \implies Q(x)$, where $P(x)$ stands for the statement $x > 2$ and $Q(x)$ stands for $x^2 > 4$. Of course, the statements $P(x)$ and $Q(x)$ contain $x$ as a free variable, and each will be true for some values of $x$ and false for others. But surely, no matter what the value of $x$ is, we would say it is true that if $x > 2$ then $x^2 > 4$, so the conditional statement $P(x) \implies Q(x)$ should be true. Thus, the truth table should be completed in such a way that no matter what value we plug in for $x$, this conditional statement comes out true.
For the case where $P(x)$ and $Q(x)$ are both false:
But now consider the case $x = 1$. Then $x < 2$ and $x^2 = 1 < 4$, so $P(x)$ and $Q(x)$ are both false, corresponding to line one in the truth table. We have tentatively placed a $T$ in this line of the truth table, and now we see that this tentative choice must be right. If we put an $F$ there, then the statement $P(x) \implies Q(x)$ would come out false in the case $x = > 1$, and we’ve already decided that it should be true for all values of $x$.
And, for the case where $P(x)$ is false and $Q(x)$ is true:
Finally, consider the case $x = −5$. Then $x < 2$, so $P(x)$ is false, but $x^2 = 25 > 4$, so $Q(x)$ is true. Thus, in this case we find ourselves in the second line of the truth table, and once again, if the conditional statement $P(x) \implies Q(x)$ is to be true in this case, we must put a $T$ in this line.
So, we have:
$$\begin{array}{|c|c|c|} \hline P&Q&P\Rightarrow Q\\ \hline F&F&\textbf{T}\\ F&T&\textbf{T}\\ T&F&F\\ T&T&T\\ \end{array}$$
Questions
I'm able to follow this example, but I think I'm missing its gist. By the text in bold in the first quote above, is Velleman saying that because the statement "if some number x is greater than 2, then its square is greater than 4" is a fact, it holds regardless of the value of x? But does it matter whether it is a fact?
I then came across this answer, which quotes Enderton's explanation for lines 1 and 2 of the truth table:
For example, we might translate the English sentence, ”If you're telling the truth then I'm a monkey's uncle,” by the formula $(V \implies M)$. We assign this formula the value $T$ whenever you are fibbing. In assigning the value $T$, we are certainly not assigning any causal connection between your veracity and any simian features of my nephews or nieces. The sentence in question is a conditional statement. It makes an assertion about my relatives provided a certain condition — that you are telling the truth — is met. Whenever that condition fails, the statement is vacuously true.
Very roughly, we can think of a conditional formula $(p \implies q)$ as expressing a promise that if a certain condition is met (viz., that p is true), then $q$ is true. If the condition $p$ turns out not to be met, then the promise stands unbroken, regardless of $q$.
I like the characterisation of a conditional as a promise, and I think this is what Velleman is trying to convey as well, but I'm not sure. Also, wouldn't this change the interpretation of the truth table? The last column now tells us whether or not a promise is kept, as opposed to whether or not a proposition is true.
