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