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I've come to find I don't understand how to prove an implication in which the consequent is a disjunction. Allow me to explain.

For a simple example, take this: "Prove: If $\frac{x}{(x-2)} \leq 3$, then $x<2$ or $x\geq 3$". Considering there's a disjunction in the consequent, does that mean I would theoretically only need to prove one of the component statements in this disjunction? I.e., would proving If $\frac{x}{(x-2)} \leq 3$, then $x<2$ suffice? A disjunction is true as long as at least one of its components is true, right? This would mean the implication has truth values $T \implies T$, and it'd be a correct implication.

However, in a recent question I asked, the above implication was solved by showing that both $x<2$ and $x\geq3$ hold. Isn't this redundant?

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    $\begingroup$ In general, it is correct; it is enough to prove one of the disjuncts. $\endgroup$ – Mauro ALLEGRANZA Oct 22 '17 at 16:08
  • $\begingroup$ Consider $n \le 0$; this is equivalent to $n < 0 \lor n =0$. Thus, if we have proved $n < 0$ we can conclude with $n \le 0$. $\endgroup$ – Mauro ALLEGRANZA Oct 22 '17 at 16:09
  • $\begingroup$ But the case above is different: the two "solutions" of the disequation are both solutions (mutually exclusive). $\endgroup$ – Mauro ALLEGRANZA Oct 22 '17 at 16:14
  • $\begingroup$ @MauroALLEGRANZA Right, but isn't the general form $P \implies (Q \lor R)$ equivalent to $(P \land \neg Q) \implies R$? So why would I have to show that both $x<2$ and $x \geq 3$ hold? $\endgroup$ – Ius Klesar Oct 22 '17 at 16:24
  • $\begingroup$ @MauroALLEGRANZA We don't know the value of x. I think that x < 2 is more like "x can take on some value less than 2" instead of "x has some value less than 2" here. It is a contradiction to say that "x has some value less than 2" and "x has some value greater than 3". It's not a contradiction to say that "x can take on some value less than 2" and "x can take on some value greater than or equal to 3" here. $\endgroup$ – Doug Spoonwood Oct 22 '17 at 18:32
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It would suffice, of course. But can you show that? Yes? Then you most definitely had an error in your reasoning.

Consider $x=3$. In this case $\frac{x}{x-2} = 3 \leq 3$ but $x<2$ does not hold. That's why you still have the second statement $x \geq 3$.

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One way to try and prove this is to prove the contrapositive.

The contrapositive of $ P \rightarrow ( Q \lor R)$ is $(\neg Q \land \neg R) \rightarrow \neg P$

Applied to your statement, that amounts to proving that:

If $2 \le x < 3$ then $\frac{x}{x-2} >3$ (or it is undefined)

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  • $\begingroup$ I think you used more than just the contrapositive there. Though I guess you didn't define 'contrapositive' and could have a different definition... $\endgroup$ – Doug Spoonwood Oct 23 '17 at 0:20
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With conditionals, I believe that the most generally applied method of proof consists in assuming the left part of the conditional and then deriving the right part of the conditional.

Now, for a conditional with a disjunction in the consequent, as Mauro's comments suggest, we first want to ascertain the nature of the disjunction. Is it an inclusive disjunction or an exclusive disjunction? In other words, can both of the disjuncts hold true or can only one of the disjuncts hold true?

If both of the disjuncts X, Y can hold true, we have some options. One option lies in proving that one of the disjuncts does hold. Consequently, the disjunction will follow. Another option starts by assuming the negation of the disjunction. Since ($\lnot$(X$\lor$Y)$\rightarrow$($\lnot$X$\land$$\lnot$Y)), ($\lnot$X$\land$$\lnot$Y) follows. Then we can consider $\lnot$X and $\lnot$Y separately in turn. If either $\lnot$X or $\lnot$Y more immediately leads to a contradiction, then so does $\lnot$(X$\lor$Y) and (X$\lor$Y) follows as true.

So, for your example suppose that $\lnot$ ((x < 2) $\lor$ (x >= 3)) under the scope of the left part of your conditional. Thus, ($\lnot$(x<2) $\land$ $\lnot$(x >= 3)). So, we can have any x such that x > 2. But, suppose that x = 2.1. Then from the left part of your conditional we obtain (2.1 / .1) <= 3. But, 21 <= 3 is false and thus we have a contradiction. Consequently, $\lnot$ ((x < 2) $\lor$ (x >= 3)) is false and ((x < 2) $\lor$ (x >= 3)) is true.

Therefore, your conditional follows.

You might not find that convincing. Well, if so or if not so, then I don't find it convincing either. But, I don't know of any logical errors. And proofs don't have to convince to work.

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