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I'm working through "How to think like a mathematician" for uni and there is this part of a question:

Rewrite the following as 'if ..., then ...' statements:
...
(d) Regular work is not necessary to pass the course.

so my first instinct is to say: "If the course is passed then regular work might have been done" but that just feels messy and I can't really translate that into symbols i.e. A implies B

I was thinking that I could exploit the nature of the implies: "If regular work is done then the course is passed" but I am not sure if that is entirely equivalent, and just cheesing the question.

There aren't any solutions in the book and the online ones don't have this section in them.

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I would write it as $¬(A\implies B)$.

Where $A$ means the course was passed and $B$ implies regular work was done.

Notice that $A\implies B$ means that regular work is needed to pass the course.

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  • $\begingroup$ PS: mathematicians don't work like that, you might be confusing mathematicians with robots. $\endgroup$ – Jorge Fernández Hidalgo Aug 28 '16 at 17:45
  • $\begingroup$ I'm doing computer science so... $\endgroup$ – Cjen1 Aug 28 '16 at 17:46
  • $\begingroup$ makes sense${}{}{}{}{}{}$ $\endgroup$ – Jorge Fernández Hidalgo Aug 28 '16 at 17:50
  • $\begingroup$ Sorry took me a while to work through your answer, truthfully this is the first day I've actually delved into logic in any formal sense $\endgroup$ – Cjen1 Aug 28 '16 at 17:54
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Let $p$ be course was passed, and $q$ be regular work was done.

The statement "regular work is not necessary to pass the course" means that knowing that the course was passed doesn't say anything about whether regular work was done.

$\neg (p \rightarrow q)$ is not equivalent to that. It says that if $p$ is true and $q$ is true (you pass the course and you did regular work) then the whole thing becomes false.

I think $p \rightarrow (q \lor \neg q$) would be a better expression.

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