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Let$\ P_0 $be our original proposition and$\ P_1 :a<b $ and$\ P_2:a<c $ the statements that are equivalent to$\ P_0 $. Now, if it is known that$\ b<c $, then there is an interval of values of$\ x $ such that$\ b<x<c $. Since both$\ P_1 $ and$\ P_2 $ are necessary and sufficient for the truth of$\ P_0 $, doesn't it mean that for the latter to be true it is only needed that$\ a $ assumes any value respectively$\ <b $ and$\ <c $, and we can't just say that$\ a$ must be$\ <b $? If so, recall $\ b<x<c $. If$\ a $ is in this interval, then$\ P_1 $ is false and $\ P_2 $ is true, leading to the contradictory result that $\ P_0 $ is at the same time true and false. Thus, may we conclude that$\ P_0 $ is false?

I hope I made myself clear.

EDIT: I'm putting here some more details from my answer to Nick and paw88789. what I'm saying is that, known that $\ P_2 $ is a necessary and sufficient condition for $\ P_0 $, one should be able to pick any $\ x<c $ for $\ P_0 $ to be true. But, as it turns out, it's not like that. In fact, if $\ b<a<c $ we have that $\ P_0 $ is both true and false, which is a contradiction, so they're all false. Alternatively, we may note that it must be $\ a<b $ to avoid the "contradictory interval". Thus, $\ P_2 $ loses its sufficiency, which is absurd since it was demonstrated. This is why I thought of concluding that all the propositions are false.

My latest guess is that since both $\ P_1 $ and $\ P_2 $ have been proven to be equivalent to$\ P_0 $, the truth of either of the three statements implies that of the others, and that's exactly because $\ a $ can't lie in the "contradictory interval", so showing that$\ a <c $ gives us that$\ a <b $ as well. That is, we can't deduce the falsity of the three statements.

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    $\begingroup$ What is this original proposition $P_0$? Are you sure that it really is equivalent to both $P_1$ and $P_2$? $\endgroup$
    – Adriano
    Sep 2, 2014 at 21:25
  • $\begingroup$ @Adriano It is an exercise that I came up with after bumping into a completely similar situation, days ago. So I can't really remember what$\ P_0 $ dealt with, but it definitely wasn't something a priori related to $\ P_1 $ and $\ P_2 $, so that doesn't really matter. I remember that its equivalence to both $\ P_1 $ and $\ P_2 $ was rigorously proved. $\endgroup$ Sep 2, 2014 at 21:38
  • $\begingroup$ If $P_1$ and $P_2$ are logically equivalent and necessary (and sufficient) conditions for $P_0$ to be true, then proving one false should be enough to prove $P_0$ false, I think. This is just my intuition - I can't prove this formally. More or less...I think I agree with you. $\endgroup$
    – 123
    Sep 2, 2014 at 23:03
  • $\begingroup$ @mathtastic Your intuition is correct, but I'm not talking about that. I've made it clearer in the edit. $\endgroup$ Sep 3, 2014 at 0:42

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You have a proposition $P_0$ and you say it is equivalent to $P_1:a<b$ and it is also equivalent to $P_2: a<c$. You say you also know $b<c$ and I assume they are all real numbers. This is not enough to conclude that these statements cannot all be true together, e.g. $(a,b,c)=(1,2,3)$ satisfies these conditions. It's different if $a$ is a free variable:

Let's look at $$S_1=\{a|a<b\}, S_2=\{a|a<c\}.$$ We know $S_1=S_2$ because $P_1$ is equivalent to $P_2$. We also know $\frac{b+c}{2}\in S_2\setminus S_1$ because $b<\frac{b+c}{2}<c$. We therefore have arrived at a contradiction. If you were attempting to do a proof by contradiction, assuming $P_1$ or $P_2$, knowing that $b<c$ and that $P_1$ is equivalent to $P_2$, then yes you can conclude that all of $P_0, P_1, P_2$ are false. Otherwise, you have derived a contradiction, hopefully a mistake or misunderstanding somewhere, otherwise we are all in trouble...

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  • $\begingroup$ actually, $b\in S_2\setminus S_1$ if $b<c$, no need for $\frac{b+c}{2}$ $\endgroup$
    – planckh
    Oct 14, 2014 at 23:35
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Nick had given an answer and it looks like it was deleted, I'm sorry if it was me, I didn't do it on purpose. He had pointed out that $\ P_1 $ and $\ P_2 $ must be equivalent as well, thus, the three statements can't be true.

Yeah, in general I would agree with him. The fact is that$\ P_1$ and$\ P_2 $ are inequalities, and it was perfectly clear that $\ b \neq c $. I know, what Nick said should always hold, yet I'm sure those inequalities were not the same... Anyway, what I'm saying is that, known that $\ P_2 $ is a necessary and sufficient condition for $\ P_0 $, one should be able to pick any $\ x<c $ for $\ P_0 $ to be true. But, as it turns out, it's not like that. In fact, if $\ b<a<c $ we have that $\ P_0 $ is both true and false, which is a contradiction, so they're all false. Alternatively, we may note that it must be $\ a<b $ to avoid the "contradictory interval". Thus, $\ P_2 $ loses its sufficiency, which is absurd since it was demonstrated. This is why I would conclude that all the propositions are false.

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One way out of your predicament would be if $a$ had to be an integer while $b$, $c$ were reals. Then for instance we could have:

$P_0: a<2$

$P_1:a<1.8$; and

$P_2:a<1.9$

These would be consistent with your problem (since $a$ is to be an integer). That is, for all $a\in\mathbb{Z}$, $P_0$ is logically equivalent to both $P_1$ and $P_2$.

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  • $\begingroup$ Yes,$\ a $ is a natural number whereas$\ b $ and$\ c $ are positive real numbers, but as I said in a comment,$\ P_0 $ is not an inequality. And this isn't an impossible imaginary situation because I bumped into it days ago, reading through various stuff, though I forgot about the specific context. My guess is that since both $\ P_1 $ and $\ P_2 $ have been proven to be equivalent to$\ P_0 $, the truth of either of the three statements implies that of the others, and that's exactly because $\ a $ can't lie in the "contradictory interval", so showing that$\ a <c $ gives us that$\ a <b $ as well. $\endgroup$ Sep 3, 2014 at 11:34
  • $\begingroup$ That is, we can't deduce the falsity of the three statements. $\endgroup$ Sep 3, 2014 at 11:37

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