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For example. Say I have a conditional

$$(\frac{x}{x-2} \leq 3 \ \land \ x \geq 2) \implies x \geq 3$$

then clearly, when $x=2$, I have a problem since then my fraction is undefined.

Now, suppose I wish to prove the above implication strictly by means of a direct proof. That means assuming the antecedent, and from there on trying to prove that it logically leads to the consequent.

However, if I assume the above antecedent $(\frac{x}{x-2} \leq 3 \ \land \ x \geq 2)$, then what do I do about the $\geq$? After all, this means $= \, or \, >$, and since I'm assuming this to be true (as part of the overarching conjunction being true), how do I deal with the conflict of both my antecedent being correct, but $x=2$ being impossible and incorrect?

Is it valid for me to say "Because in the case of x=2, the fraction is undefined, our assumption reduces to $(\frac{x}{x-2} \leq 3 \ \land \ x > 2)$" and then go on to prove $(\frac{x}{x-2} \leq 3 \ \land \ x > 2)$?

That somehow feels like cheating, or at the least like giving some sort of vacuous/irrelevant proof because I've reduced my assumption (and, by extension, my implication) to something it strictly wasn't. Can somebody shine some light on this?

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  • $\begingroup$ Genuine question: do we say the antecedent is false when $x=2$? I think assigning a truth value to an inequality where one side is undefined could be problematic... Of course, if we can say the antecedent is false when $x=2$, surely you can ignore this case when doing a direct proof, which entails assuming the antecedent is true. $\endgroup$ – Theoretical Economist Oct 25 '17 at 18:51
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I think that, since the quantification on $x$ is implicit, you can assume that the $x$ concerned are those for which the assertions $\frac{x}{x-2}\leq 3$ and $x\geq 2$ make sense simultaneously, and what you are trying to prove is more precisely $$\forall x\in\mathbb{R}\setminus\{2\},\frac{x}{x-2}\leq 3\,\wedge x\geq 2\implies x\geq 3.$$

It reminds me a question I asked of the same nature (here), with a nice answer from John M. Lee.

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  • $\begingroup$ Quantification is only implicit if the variables in an implication are not explicitly bounded, and then the convention is to implicitly quantify the variable over all real numbers. This is not the case here. I don't think you can just play around with domains like that, and implicitly quantify it over a set which doesn't contain a specific element to make your proof easier. $\endgroup$ – Ius Klesar Oct 25 '17 at 18:45
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    $\begingroup$ Perhaps my answer is too naive for you (I am not a logician) : here I just say that if I have two assertions $P(x)$ and $Q(x)$ which have sense on sets $A$ and $B$, when I write the new assertion $P(x)\implies Q(x)$ I implicitely write $\forall x\in(A\cap B), P(x)\implies Q(x)$. I just take as definition for assertion "a sentence which makes sense and is either true, either false", so I would easily understand if you are looking for a more logical-technical answer! $\endgroup$ – Balloon Oct 25 '17 at 18:58
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    $\begingroup$ It's not true that "the convention is to implicitly quantify the variable over all real numbers". In practice it's like what we teach: the domain of $f(x)=\frac{x^2-1}{x-1}$ is implicitly the subset of the real numbers for which evaluation is defined, namely $\mathbb{R} - \{1\}$, even if there is a trick for making sense of the case $x=1$. Similarly, your conditional is a Boolean valued function, and its domain is implicitly the subset of the real numbers for which the evaluation is defined, namely $\mathbb{R} - \{2\}$, even if there is a trick for making sense of the case $x=2$. $\endgroup$ – Lee Mosher Oct 25 '17 at 20:49
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    $\begingroup$ And, by the way, I like that answer by @JohnLee quite a lot. $\endgroup$ – Lee Mosher Oct 25 '17 at 20:51
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You said $(\dfrac{x}{x-2}\leq 3~~ \wedge ~~x\geq 2 )\Rightarrow x\geq 3$. So when you have the antecedent, you get the consequent. But, the thing is that you just have the first inequality of the antecedent if $x \not =2$, if $x=2$ you can't say anything about $\dfrac{x}{x-2}$, because it is not defined as a real number, so you can't compare it with the $\leq$ of $\Bbb{R}$. Since $x \geq2$ iff ($x>2 ~~\vee~~x=2$) there is no problem with the first inequality being satisfied just when $x \not = 2$, because the second inequaliuty can still be satisfied (since $\vee$ requires just one to be satisfied). Therefore, there is no problem with the implication.

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  • $\begingroup$ This is not correct. An implication of the form $(P \land Q) \implies R$ is tautologically equivalent to $P \implies (Q \implies R)$ which, if $Q$ consists of a conjunction $Q_1 \lor Q_2$ like in the implication in the original post, we could write it as $P \implies [(Q_1 \lor Q_2) \implies R]$, which in turn is equivalent to $P \implies [(Q_1 \implies R) \land (Q_2 \implies R)]$. In other words, both implications $x>2 \implies x \geq 3$ AND $x>2 \implies x = 3$ should be proven for the entire implication to hold. Your suggestion of only proving one therefore seems wrong. $\endgroup$ – Ius Klesar Oct 25 '17 at 19:23
  • $\begingroup$ You meant $... AND ~~ x=2 \Rightarrow x \geq 3$, right? $\endgroup$ – Robson Oct 25 '17 at 20:13
  • $\begingroup$ Ok, you can see that $P \Rightarrow (Q_1 \Rightarrow R)$... but $P \Rightarrow (Q_2 \Rightarrow R)$ is also true because this is equivalente to $P \Rightarrow (\neg Q_2 \vee R)$ Since we have, in this case, $P \Rightarrow \neg Q_2 $, the truth follows. $\endgroup$ – Robson Oct 25 '17 at 20:21

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