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Let $been(x,y)$ denote "person x has been to place y."

Express “Every person has been to 2 or more places” using quantifiers and $been(x, y)$.

This one is completely stumping me. I'm sure that I haven't come across this kind of concept before, and can't wrap my head around how I would approach it. I suspect the quantifiers should be $∀x$, $∀y$, and that it may involve negating the statement "every person has been to one or zero places," but that's as far as I can manage to go. Any help greatly appreciated.

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  • $\begingroup$ Negating "Every person has been to zero or one places" gives "There exists a person who's been to at least two places" $\endgroup$
    – Arthur
    Jan 19 '17 at 23:52
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    $\begingroup$ Ah, okay, you're right. If I were to follow that path, it should be the negation of "At least one person has been to one or zero places." $\endgroup$
    – Mock
    Jan 19 '17 at 23:56
  • $\begingroup$ Yes. Actually, in my set theory book, the symbol $\exists x$ is defined as $\lnot\forall x\lnot$. Other books might do it differently, but nonetheless, it's a generally accepted fact that those two mean the same. $\endgroup$
    – Arthur
    Jan 19 '17 at 23:57
  • $\begingroup$ math.stackexchange.com/questions/2103616/… This page has the same question. I found the comments on the Op's question most helpful. The question was a surprise to me too. $\endgroup$
    – user400188
    Jan 20 '17 at 2:56
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Try this one: $$ \forall x \in \text{Persons}(\exists y_1, y_2\in \text{Places}(y_1 \neq y_2 \land \text{been}(x, y_1)\land \text{been}(x, y_2)) $$

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  • $\begingroup$ That's a slightly different symbol convention than my course uses, but based on this concept I stated "$∀x, ∀a, ∀b$, $been(x,a)$^$been(x,b)$^$(a≠b)$." I believe I'm stating the same thing? $\endgroup$
    – Mock
    Jan 20 '17 at 0:03
  • $\begingroup$ First off, \land (Logic AND) makes the $\land$ symbol. Second, I think you need the $a$ and $b$ to be quantified by $\exists$, not $\forall$. Otherwise, yes. $\endgroup$
    – Arthur
    Jan 20 '17 at 0:05
  • $\begingroup$ I don't follow why $a$ and $b$ would be quantified by $∃$? The $∀$ says "take any 2 places from the list of all available places," and $a$≠$b$ is specified later. Wouldn't $∃a,$ $∃b$ leave the possibility that some combination of places is NOT valid? Which contradicts the statement. $\endgroup$
    – Mock
    Jan 20 '17 at 0:19
  • $\begingroup$ No. If you qualify them with $\forall$, then what you're saying is that "For any person, and any two places, that person has been to those two places, and those two places are not the same". It can be roughly translated to "no place is equal to any other (not even itself), and everyone has been everywhere". If you qualify with $\exists$, you're saying "For every person, there exists two places that are not the same, both of which the person has been to", which is what you want. $\endgroup$
    – Arthur
    Jan 20 '17 at 0:21
  • $\begingroup$ You're right, my mistake. Marking this as solved, thanks for the help! $\endgroup$
    – Mock
    Jan 20 '17 at 0:25

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