Say I have the four following logical statements, all over the domain of all integers.

  1. (∀a,∀b)[a>b]
  2. (∀a,∃b)[a>b]
  3. (∃a,∀b)[a>b]
  4. (∃a,∃b)[a>b]

I feel like they're all asking practically similar things, but I'm getting on confused on what exactly for all means. I'm writing what I think each statement is asserting literally, please correct me if I'm wrong:

  1. All integers are greater than each other? (This is the one I'm struggling the most with)
  2. There is an integer b that is less than all other integers
  3. There is an integer a that is greater than all other integers
  4. There is an integer a that is greater than an integer b

So if this is the case, I'm assuming that all are false except for (4). But in the event that I have correctly understood all four of these statements, how would you express something like how for all known integers, there is another integer that is greater and/or lesser than it?

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    $\begingroup$ (2) Is instead read as "For all $a$ there exists a $b$ such that $a>b$" In an attempt to reword it in more natural English it is saying that if Adam chooses an integer, regardless which one he chooses his friend Bill can look at Adam's choice and then with that knowledge pick an integer which is smaller than Adam's choice. For example, if Adam chose $50$ as his integer then Bill can come along and choose a smaller integer such as $49$. $\endgroup$ – JMoravitz Feb 23 at 3:21
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    $\begingroup$ I would translate 1 like "for every arbitrary pair of integers, the first one is always bigger than the second" $\endgroup$ – kimchi lover Feb 23 at 3:22
  • $\begingroup$ @JMoravitz In that case, would it be possible to write a logical statement asserting what my original interpretation was (even though it's untrue)? $\endgroup$ – user2709168 Feb 23 at 3:37
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    $\begingroup$ The statement "There is an integer $b$ that is less than all other integers" can be written as $(\exists b~\forall a)[a>b]$. Note the order is different, $\exists b~\forall a$ has different meaning than $\forall a~\exists b$. $\endgroup$ – JMoravitz Feb 23 at 3:42
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    $\begingroup$ This answer is about the same question: ://math.stackexchange.com/a/1130755/25554 $\endgroup$ – MJD Feb 23 at 4:13

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