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In negating the following statement:

Original "If the lines are not parallel, then they intersect"

My textbook says it should be "the lines are parallel and they do not intersect", but I'm not so sure.

$$p : \text{the lines are parallel}$$

$$q : \text{the lines intersect}$$

$$\neg p \rightarrow q$$

Negation: "If the lines are not parallel, then the lines do not intersect."i.e. $$\neg p \rightarrow \neg q$$

The textbook claims that the negation is "the lines are parallel and they do not intersect" i.e.

$$p \wedge \neg q$$

Unfortunately, these two are not logically equivalent.

Who's right, and why?

Scott Jodoin
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    @TheSilverDoe: You are wrong too....... – user21820 Oct 22 '20 at 20:03
  • But someone upvoted my comment so I am happy. Thanks for pointing this stupidity :) – TheSilverDoe Oct 22 '20 at 20:06
  • Couple days ago, here, as an answer to the question "How do you write a negation of the given statement?" someone replied that you put $\neg$ in front of the expression. Well, that's it, after that you just need to work your way through appropriate logical equivalences to get to the desired form. – anfauglit Oct 23 '20 at 05:57
  • The piece that threw me off is that some of the informal examples in my textbook suggested that the negation of $p \rightarrow q$ is $p \rightarrow \neg q$. This is only the case when p = 1 and q = 1. For a proper 'negation', it needs to be as you said. That makes sense! – Scott Jodoin Oct 24 '20 at 15:42

2 Answers2

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None of you are right. $\neg p\implies q$ is equivalent to $p\vee q$, so its negation is $\neg p\wedge\neg q$ i.e. "the lines are not parallel and they don't intersect".

Shubham Johri
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  • I thought the negation of a conditional $$p \rightarrow q$$ is $$p \rightarrow \neg q$$ ? What am I missing here? (https://math.stackexchange.com/questions/2417770/what-is-the-negation-of-the-implication-statement) – Scott Jodoin Oct 22 '20 at 20:00
  • That is incorrect. Both $p\implies q$ and $p\implies\neg q$ are true when $p$ is false, which can't be the case with negations. – Shubham Johri Oct 22 '20 at 20:01
  • @Scott - the question you linked indicates $\lnot (p \to q) = p \land \lnot q$. OP is confused why what you're referring to is incorrect. – HallaSurvivor Oct 22 '20 at 20:02
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    @ScottJodoin The post you linked says the negation of $p\implies q$ is $p\color{red}\wedge\neg q$, not $p\implies\neg q$, which is consistent with what I wrote. – Shubham Johri Oct 22 '20 at 20:04
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    Oops! Thanks for pointing that out @HallaSurvivor. – Scott Jodoin Oct 22 '20 at 20:09
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    No worries - it happens to all of us – HallaSurvivor Oct 22 '20 at 20:09
  • @ScottJodoin I think the statements $p\wedge\neg q$ (read as "$p$ and not $q$") and $p\implies\neg q$ (read as "$p$ implies not $q$") can be easily confused... – Shubham Johri Oct 22 '20 at 20:13
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To negate an implication, it is useful to remember that it is an abbreviation.

$$p \to q = (\lnot p) \lor q.$$

A constructivist might argue that $\lnot (p \land \lnot q)$ is a better definition, but that's a subtle point. These definitions are equivalent in classical logic.

In particular, the negation of $p \to q$ is not $p \to \lnot q$, as you're claiming. If you work through the negation using either of the equivalences above, you'll find that the negation of $p \to q$ is actually $p \land \lnot q$. I'll leave it to you to apply this reasoning to your particular problem.

I hope this helps ^_^

HallaSurvivor
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