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for example : If you show up for work monday morning then you get the job,

the only cases where we know the truth of this statement is if you show up on monday morning,if you don't you can't possibly know if you would have gotten the job. Why in this case do we just assume it to be true? that seems really arbitrary and "unmathsy".

Why can't we assume that they are false by default instead?

does it being automatically assumed to be true factor in later? I have to apologise since I haven't read that far yet.

arshdixit
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    Who says we assume that to be true? – lulu May 25 '21 at 00:00
  • No, the antecedent is not necessarily assumed to be true. In some sense it is arbitrary. An example I came across (I wish I could remember the OP) when I was first learning this was the following: Suppose your professor tells you "If you get 50%, I will give you an A". Now suppose you get 40% and receive an "F". Was your professor unfair? – masiewpao May 25 '21 at 00:04
  • discrete maths with applications Susan S epp, page 40, it says that the cases where if you don't go to work on monday morning, the statement is vacuously true/true by default if you don't show up on monday morning.

    the only time when it's false is in the truth table if you show up on monday morning and you don't get the job, that's why I say it's true by default.

     – arshdixit May 25 '21 at 00:05
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    The implication may be true, but that says nothing about the hypothesis or the conclusion. – JMoravitz May 25 '21 at 00:05
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    Suppose that the implication "If you show up monday you get the job" holds. Us not showing up monday and not getting the job is perfectly acceptable. We didn't show up so us making the hiring manager angry is an acceptable outcome. It is also possible that we didn't show up monday and the hiring manager couldn't find anyone to do the work and so called us and rescheduled our interview and we get the job over the phone or something. The only "unfair" outcome that we don't expect is if we actually show up but aren't offered the job. – JMoravitz May 25 '21 at 00:09
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    Does this answer your question? In classical logic, why is $(p\Rightarrow q)$ True if both $p$ and $q$ are False? Alternatively, In classical logic, why is $(p\Rightarrow q)$ True if $p$ is False and $q$ is True? – JMoravitz May 25 '21 at 00:10
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    You may also like to read the post Assumed True until proven False. The CuriousCase of the Vacuous Truth. The punchline again is that if we know that $p\implies q$ then knowing $p$ is true is enough to know that $q$ is true but knowing $p$ is false it is not enough to know anything about $q$. It could either be the case that $q$ is false or it could also have been the case that $q$ was true. – JMoravitz May 25 '21 at 00:16

2 Answers2

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In Classical Logic, a conditional statement is basically a promise of: if this is true, then that will be true too.

No claim was made about that when this is not true, so the promise can be said to still hold in that case.

The only case where the promise is broken is when that is true but this is false.

So the promise holds when: this is false or that is true.

Graham Kemp
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A conditional, or any statement for that matter, can be considered to be true when it's "not false". Now, in your example, if you never even show up for work on Monday, there's no way that this conditional can become false because it's falsity completely hinges on you showing up. So, it has to be true.

"It's true because it's not false."

Hope this helps!

Gaurav Chandan
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  • @arshdixit To be clear, the OP is really asking why the consequent/conclusion is assumed to be true whenever the antecedent/supposition/hypothesis/premise is known to be false. However, both the OP and yourself are incorrectly stating that the (entire) conditional statement (the implication) is being assumed true whenever the supposition is known to be false. – ryang May 25 '21 at 02:15