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M: The truth table of p→q says that if p is false and then p→q is true. The author of the books goes on to clarify the doubt people have about the above statement, he uses the following scenario to explain why M is true,

"Perhaps you are bothered by the fact that P ⇒ Q is true in the last two lines of this table. Here’s an example that should convince you that the table is correct. Suppose you have just taken a test, and you ask your professor if it will be graded by the next class. Your professor makes the following promise: If you come to class, then you will get your test. Your professor is making the promise (You come to class) ⇒ (You get your test)."

He goes on to say that if you don't come to class and (get your paper)/(do not get your paper), then he din't lie, if you don't lie you say the truth, therefore he din't lie. But how can we be sure he dint lie?, the promise wasn't broken, sure, but we are not sure that it wouldn't have been broken if we had previously decided to go to class instead of not going.

ps; I haven't taken any logic courses.

Moin
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    Mathematics isn't about intentions. It is provably impossible to show that he lied, which automatically (unless you go deeply into mathematical logic and philosophy) means that he didn't. But I feel that a true answer to your question will have to go into that logic and philosophy, so this is a comment. – Arthur Dec 07 '19 at 15:45
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    See Vacuous truth, also this post. – Ethan Dec 07 '19 at 15:47
  • Every time you went to class (all 0 times!), you got your paper. Therefore the implication holds. – Zubin Mukerjee Dec 07 '19 at 15:50
  • Even if we ignore the mathematical aspect, a statement of the form "If $A$ , then $B$" does tell us nothing about the case that $A$ is false. – Peter Dec 07 '19 at 15:51
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    These vacuous truths are, in my mind, related to the empty product being equal to 1, and the empty sum being equal to 0. – Zubin Mukerjee Dec 07 '19 at 15:51
  • @Arthur Lets assume that it is impossible to prove he lied, but there isn't any reason to believe that he din't. – Moin Dec 07 '19 at 16:11
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    @MJD his question is fundamentally different from mine, and I don't have the same doubt. – Moin Dec 07 '19 at 16:17
  • It doesn't matter what he intended. He kept his promise. Thus the promise was true. – Arthur Dec 07 '19 at 16:27

2 Answers2

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$p \Rightarrow q$ by definition means $\lnot p \lor q$. Everyday language can give an intuition of the reason behind this: "if they don't discount it, I won't buy it" and somewhat equivalently "either they'll discount it or I won't buy it". A line in a truth table is to be interpreted as a disjunction of conjunctions of elementary propositions. So, if you look at the truth table for implication: $$\begin{array}{lcccc} p & 1 & 1 & 0 & 0 \\ q & 1 & 0 & 1 & 0 \\ p \Rightarrow q & 1 & 0 & 1 & 1 \end{array}$$ One way of reading it is "(p and q) or (not p and q) or (not p and not q)", which is equivalent to "q or not p". In other words, you have to take a holistic reading of the whole line, which is your definition. It doesn't make sense to say, for example, "if p is false, then it is true that p implies q".

Papa Smurf
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When we say that a formula (about the real numbers, say) such as "If $x < 5$, then $x < 10$" is true, that means, it is true for ALL values of the variable $x$, including, for example, $x=12$ (F implies F), $x=8$ (F implies T), and $x=2$ (T implies T).

This is indeed a convention of mathematical language/notation, but a very convenient one, because we'd like the interpretation of "Formula $A(x)$ is true" to be the same for implications $A(x)$ such as the one above as for e.g. $A(x)$ being $(x^2-1)=(x+1)(x-1)$, namely that the formula holds for all values of $x$.

Ned
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