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I apologize if this has been answered elsewhere, but I've run across this problem in proving inequalities a lot, and I haven't been able to find an answer so I would like to ask if the proposition is true. I feel like it is, but I'm not certain because I can't recall a time I've actually seen it used in a proof. When we say $x \leq y$, we say "$x$ is less than OR equal to $y$," so if we know $x < y$, is it enough to conclude that $x \leq y$, since one of the cases is always satisfied? Maybe the more general question is, if we ask the question is "$x, y$ or $z$", and we know it's one of them but not the other, is it enough to conclude $x$ is $y$ or $z$, if it's one but never the other?

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    $\begingroup$ $P\implies P\vee Q$ $\endgroup$ – Kapil Oct 31 '18 at 3:06
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Yes, indeed, since $x\leq y$ cannot be false when $x<y$ is true, therefore the conditional, $(x<y) \to (x\leq y)$, is held to be true.

If $x$ is less than $y$, then $x$ is less than or equal to $y$.


Essentially, $\rm P\to(P\vee Q)$ is a tautology.   A disjunction is true when at least one of its disjuncts is true.   So if a preposition is true, then its disjunction with something else is true.

I feel like it is, but I'm not certain because I can't recall a time I've actually seen it used in a proof.

I'm quite sure you have, in some form or other.   The rule of inference known as weakening, or disjunction introduction, is used quite frequently.

Well, okay, here the disjunction is hidden inside the definition for $\leq$.   But you still realised it was there.

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  • $\begingroup$ Thank you so much for the detailed explanation :) $\endgroup$ – Stawbewwy Oct 31 '18 at 5:11

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