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Domain of Discourse: People on vacation and their belongings

Vacationer(x): x is a person on vacation

Belonging(x): x is the belonging of some person

Packed(x, y): x is a vacationer that packed belonging y

Identical(x, y): true if and only if x and y are identical belongings

We can also assume "=" operator that is true when x and y are the same people or same object.

(a) Everything that was packed by someone is a belonging.

(b) Kim packed two identical things but nothing Kanye packed is identical to either of those. (Use the constants “Kim” and “Kanye” to refer to those two vacationers.)

(c) No two vacationers packed anything identical.

(d) Some vacationer packed only two things.

I am having trouble going about how to solve these four propositions. The goal is translate them into predicate logic, but I can't seem to see how to relate those predicates to solve the propositions. Like when it says "packed by someone, is a belonging" but the packed predicate defined also checks if that someone is a vacationer.

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  • $\begingroup$ What have you tried ? $\endgroup$ – Mauro ALLEGRANZA Oct 9 '19 at 6:30
  • $\begingroup$ @MauroALLEGRANZA So for part A, I am having trouble figuring out how to extract the fact vacationer(x) aspect of the packed(x) predicate. For part B, I am eyeing the word "things", which means that they're not people, but I can't figure out how to show that there aren't any people. For C, I can see how the packed predicate can fit since it shows that there is a vacationer that packed a belonging, but that evaluates to true if it's a belonging only. Part C, I am lost at the word "only" since I don't see how to fit in the numerical aspect of it. $\endgroup$ – uwuwuuwuwuw Oct 9 '19 at 7:35
  • $\begingroup$ In other words, I am completely lost and/or struggling. $\endgroup$ – uwuwuuwuwuw Oct 9 '19 at 7:36
  • $\begingroup$ Start Step-by-Step from the simpler ones : (a) Everything that was packed by someone is a belonging. For "Everything that was packed by someone" we must have $\forall x \exists y \text {Packed}(x,y)$. Thus the complete sentence will be : $\forall x (\exists y \text {Packed}(x,y) \to \text {Belonging}(x))$ $\endgroup$ – Mauro ALLEGRANZA Oct 9 '19 at 7:48
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(a) Everything that was packed by someone is a belonging.

Since the domain is both vacationers and belongings, it makes sense to assume that the 'someone' who did the packing has to be one of the vacationers. So, I would paraphrase this as:

(a)' Everything that was packed by a vacationer is a belonging.

Which can be read as:

(a)'' For everything" if it was packed by a vacationer, then is a belonging.

And thus formalized as:

$\forall x (\exists y ((Vacationer(y) \land Packed(x,y)) \to Belonging(x))$

d) to say that some vacationer packed only (and I take that as exactly) two things, is to say that there are two (different) things that the vacationer packed, but no third. So:

$\exists x (Vacationer(x) \land \exists y \exists z (y \neq z \land Packed(x,y) \land Packed(x,z) \land \neg \exists w (w \neq y \land w \neq z \land Packed(x,w)))$

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  • $\begingroup$ How would I go about part d in specifying there are are no more than 2 things being packed? $\endgroup$ – uwuwuuwuwuw Oct 10 '19 at 1:42
  • $\begingroup$ @questionsfromuw I added that one to my Answer $\endgroup$ – Bram28 Oct 10 '19 at 2:16

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