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New to both math.stackexchange, please let me know if I am breaking any rules with this question. Also, new to logic, so sorry if this is a stupid question.

I am trying to get my head around Quantifiers. Say you have a statement that can be expressed as:

$$\exists x (Rx \rightarrow \forall y Ry)$$

If the statement is written in such a way that $\forall yRy$ could be both True and False, would this statement be both true and false, and if not, how do you determine the truth of it?

For example, with "there is someone who is eating such that if they are eating then everyone else is eating", $Rx$ will always be true, but will $Ry$ both true and false as we do not know if the other people are actually eating?

GalacticPonderer
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1 Answers1

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OK so you are pretty close on this.

Imagine a society in which, when the queen eats, everybody else then eats. She would be that $x$ there could be other such people (e.g. the king) and still the statement would be true. Are all the people actually eating? Who knows? The statement just says that there is a person such that when they eat, everybody eats. It's not true of our society but it may be true of some (in practice they would have to consist of relatively few people :)).

OK so having nosed around the previous similar question, I now realise this is true in all cases and can be proved by pure logic. It goes like this, if $Ry$ is true for all $y$ then we can choose any $x$ as $P \rightarrow \top$ is true for all propositions $P$. If there is an $x$ for which $Rx$ is false then choose $x$ to be that. In this case, $Ry$ is not true for all $y$, neither is $Rx$ and so we have $\bot \rightarrow \bot$ which is true. In either case, the statement is true!

Simon Terrington
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  • Thanks. To clarify, and using your example, if the queen is eating, it must be true that the other people must also be eating? How could you prove this, as surely you cannot know if everyone is eating at the same time as the queen or not, or is this just a side effect of the logical statement? (i.e. what I am not getting) – GalacticPonderer Nov 08 '20 at 17:37
  • OK so of course proof in practical situations are very difficult. It's hard to think of mathematical examples that would fit this statement but how about this? If one number is a finite distance from the number $0$ then all counting numbers are. Let's choose one number that is, say $1$. Now consider any other number ($n$), this is $n+1$ away from $0$ which is always finite. This is our proof. – Simon Terrington Nov 08 '20 at 19:34