How is "p implies q" same as "q unless not p"?

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I want to know how is "p implies q" same as "q unless not p"?
ie how is "$p\Rightarrow q$" same as "$q$ unless $\neg p$" ?

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First, "unless" is not formal mathematical language. It is loose terminology.

Read it longer as "$q$ is true unless $p$ is false." This is read as "either $q$ is true or $p$ is not true."

If this is true, and $p$ is true, it in necessarily the case that $q$ is true.

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It doesn't!

"p implies q", $p\to q$ is equivalent to "q or not p", $q\vee \neg p$.   This is an inclusive or; it does not exclude the possibility that $q$ and $\neg p$ may be both true.

That is not the same thing as "q unless not p", which would be an exclusive or.

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Suppose: "If $p$, then $q$."

There are two cases: $p$ can be true, or it can be false. If $p$ is true, then so is $q$. So it follows that $q$ holds unless $p$ is false. Ergo, $q$ unless not $p$.

Now suppose: "$q$ unless $\neg p$."

We're trying to prove "If $p$, then $q$." So assume $p$. Since its not the case that $\neg p$ holds, hence from "$q$ unless $\neg p$" we deduce that $q$ holds. Ergo, if $p$, then $q$.

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Let's take an example. You can find the same in Rosen's book

Conditional statement: if she studies math (p), then she will find a good job (q)

This is of the type: If p, then q.

q unless p: She finds a good job unless she does not study math.

Here q is true only is ~p (not p) is false. If ~p is true then q will become false.

Hope this helped!

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Sorry, but I disagree with the last example (given by Aditya Raj). I would interpret "She finds a good job unless she does not study math" as saying that the only circumstance that would cause her not to find a good job is failing to study math. But this is not at all implied by the original conditional statement which does guarantee that studying math will lead to a good job, but allows for other possibilities (does not make it a requirement).