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Having a bit of difficulty understanding the conditional ($\rightarrow$) in mathematical logic. I read up on the already-existing questions and it did help me understand it better (the 'promise' analogy really helps!).

But then I continued reading my text-book, which contains something that muddled me up a bit, again. Here it is:

Consider the following conditionals.
(i) If $x$ is an odd integer, then 4 divides $x^2 - 1$.
(ii) If $x$ is an odd integer, then 4 does not divide $x^2 - 1$.
(iii) If $x$ is not an odd integer, then 4 divides $x^2 - 1$.
(iv) If $x$ is not an odd integer, then 4 does not divide $x^2 - 1$.

You certainly know that 4 divides $x^2 - 1$, if $x$ is an odd integer. You will easily see that (ii) is false, while the rest are true, because in all the three statements conclusion is a fact.

Okay, so I understand why (i) is true, and (ii) is false.

Then I cross-checked (iii).
At first I thought that if $x$ isn't an odd integer, then 4 doesn't divide $x^2 -1$, so it couldn't be true. After which I realised that $x$ could be $\sqrt{13}$ and satisfy both conditions.

But then how is (iv) true? If x isn't an odd integer, it could be $\sqrt{13}$. In that case 4 would divide $x^2 - 1$. That's what's confusing me.

$P.S:$ There's also the possibility that the example isn't a good one. Is it a good idea to have both statements related to each other? (As they are in this case)

Shaun
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mikhailcazi
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    Each of these four conditionals don't have a truth value until you plug in a value for $x$. The textbook most probably supposes you to imagine an implicit "For every integer $x$ it holds that ..." in front of each of them. – hmakholm left over Monica Jan 09 '14 at 10:51
  • (responding to my own comment above) ... or does it? If it claims that (iii) and (iv) are both true, it must be because it has already decided on a value for $x$ which happens to be an odd integer. But "because in all three statements conclusion is a fact" makes no sense at all here. – hmakholm left over Monica Jan 09 '14 at 10:57
  • @MauroALLEGRANZA: That's only right under the interpretation where $p$ and $q$ both happen to be true. – hmakholm left over Monica Jan 09 '14 at 10:58
  • @HenningMakholm If they implied a "For every integer x it holds that..." then how could (iii) be true? That's what confused me in the first place. If x (being an integer) isn't an odd integer (which means it's an even integer), then how would 4 divide $x^2 - 1$? – mikhailcazi Jan 09 '14 at 11:14
  • @mikhailcazi: It can't. After I read your quote closer, it seems just to be unsalvageably wrong. – hmakholm left over Monica Jan 09 '14 at 11:22
  • @MauroALLEGRANZA I don't think these are examples of conditional logic (they introduce the concept of $\rightarrow$ after this particular paragraph) - but just 'pre-examples' to give us an intuitive look at the concept. However, instead of helping me understand $\rightarrow$ it basically just confused me. – mikhailcazi Jan 09 '14 at 11:36
  • @HenningMakholm So would you say "Bad example"? :/ – mikhailcazi Jan 09 '14 at 11:38
  • @HenningMakholm Aaaand, if you don't take the case of $x \epsilon$ I, then 4 can't be true! – mikhailcazi Jan 09 '14 at 11:39

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For the correct explanation of this passage, see Peter Smith's answer to this post.

The discussion dates back to :

  • A.N.Whitehead & B.Russell, Principia Mathematica (1st ed - 1910) : pages 20-21 of the Introduction.

The context is the elucidation of formal implication [in modern symbols : $\forall x (\phi (x) \rightarrow \psi (x))$ ]:

A formal implication states that, for all possible values of $x$, if the hypothesis $\phi(x)$ is true, the conclusion $\psi(x)$ is true. Since " $\phi (x) \rightarrow \psi (x)$ " will always be true when $\phi (x)$ is false, it is only the values of $x$ that make $\phi (x)$ true that are important in a formal implication; what is effectively stated is that, for all these values, $\psi (x)$ is true.

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Mauro ALLEGRANZA
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    The truth table does not DEFINE implication. $\rightarrow$ is defined by its inference rules, and the truth table is (an awful) shorthand of encoding those when the law of the excluded middle holds. The definition of implication is that its truth is equivalent to our ability to derive the truth of $A$ from the truth of $B$. – Vladimir Sotirov Jan 09 '14 at 21:45
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    @Vladimir: The symbol $\to$ is just ink; it doesn't need any definition. Its meaning is a matter of semantics, and the semantics of connectives in classical logic is defined by truth tables. Once you choose a proof system, its rules determine what you can do with $\to$ in that particular proof system, but the symbol itself or its meaning is unaffected by that. – hmakholm left over Monica Jan 09 '14 at 22:05
  • I am aware of the ink, when I wrote "$\rightarrow$" I meant "implication"; sorry if I was unclear. – Vladimir Sotirov Jan 09 '14 at 22:10