0
$\begingroup$

I think I know how to form sentences in statement logic if it's an "if statement" like (A) and (B) below, but how do I express adjective like "not so easy" or imperative like "Choose X or Y", as shown in (C) and (D) below?

(A) If Lionel Messi makes another goal, then he will make a hat trick and Barcelona will win the match" 
    “Lionel Messi makes another goal” → “he will make a hat trick” Ʌ “Barcelona will win the match” 

(B) If Musikhjälpen 2019 gains over 11 million Euro, then it becomes a record, otherwise Musikhjälpen 2018 holds the record. 
    “Musikhjälpen 2019 gains over 11 million Euro” → “2019 becomes a record”
    “Musikhjälpen 2019 doesn't gain over 11 million Euro” → “2018 holds the record”

(C) It is not so easy to get both a job that is interesting and a job with a high salary even if you search many companies and it is easy to just find a job itself. 
      "??????"

(D) Choose a ferry trip with a cabin or a flight and hotel, but not both of them
(“a ferry trip with a cabin” V “a flight and hotel”) Ʌ ¬(“ferry with a cabin” Ʌ “flight and hotel”)
$\endgroup$
1
$\begingroup$

It's a little bit unclear what the person who has written these exercises is expecting of you, but here is one way that these might be reasonably interpreted:

It seems for (C) that you need to have a partial ordering over the set of potential jobs that tells you whether one jobs is easier to obtain than another.

If we call $J$ the complete set of "jobs", then we can define a partial ordering relation over those jobs that orders them by how difficult they are to obtain by constructing $R \subseteq J\times J$ and saying that for $a,b\in J$, "$a$ is at least as easy to obtain as $b$ iff $(a,b)\in R$."

You may then construct some predicates on $J$, such as $\mathrm{isInteresting} : J \to \mathrm{Boolean}$ and $\mathrm{hasHighSalary} : J \to \mathrm{Boolean}$. (Predicates are in general the usual way one might handle something like "adjectives" logically.)

Then the statement of (C) is just that $\forall a,b \in J, \mathrm{isInteresting}(b) \land \mathrm{hasHighSalary}(b) \implies (b,a) \not\in R$

For question (D), I think the purpose of the imperative is to have you thinking about the solution as involving applying the constraints to a particular element. If you think about a type (or set) of all possible travel options, which we might call $T$, then you could perhaps phrase this by asserting the existence of a particular choice of travel option and given certain constraints, perhaps as $\exists c\in T, c=\textrm{``ferry trip with cabin''} \lor c=\textrm{``flight and hotel''}\land \neg (c=\textrm{``ferry trip with cabin''} \land c=\textrm{``flight and hotel''})$

Although without knowing the context that you're studying this in, I can't predict for certain that this is what is expected of you.

$\endgroup$
  • $\begingroup$ Note that instead of using the more common set-theoretical definition of a partial ordering as I described above, you could also define it as a predicate on two arguments with something like $ \mathrm{isAtLeastAsEasyAs}: J\times J \to \mathrm{Boolean}$ and then replace $(b,a)\not\in R$ with $\neg \mathrm{isAtLeastAsEasyAs}(a,b)$. There are many ways you could do something like this. $\endgroup$ – Jack Crawford May 19 at 12:50
  • 1
    $\begingroup$ Constructing predicates like hasHighSalary, isInteresting in "job statement" and adding one more variable like c in "trip statement" totally make sense. This is exactly what I was looking for. You interpreted really excellent and that was quite helpful. Thank you very much! $\endgroup$ – Shinichi Takagi May 19 at 17:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.