How is "so" correctly translated into predicate logic? I have come across an exercise that asks to have “There is only one ball, so you need to have it” translated into predicate logic. Using the predicates Ball(x) for x is a ball and Have(x, y) for x must have y, I translated it this way:
$$\exists x((Ball(x) \land \forall y(Ball(y) \rightarrow (x = y))) \rightarrow Have(you, x))$$
The answer in the back, on the other hand, is:
$$\exists x(Ball(x) \land \forall y(Ball(y) \rightarrow (x = y)) \land Have(you, x))$$
I want to know whether I am wrong or not, and why. The exercise and answer can be found here
https://textbooks.open.tudelft.nl/textbooks/catalog/view/53/144/370-1, on page 51 (exercise 8) and page 189 respectively.
Thank you.
 A: I like neither translation.
Yours is certainly not right, since your statement would be rather vacuously true if there exists something that is not a ball: that something would make $Ball(x)$ False, hence the whole antecedent of the conditional false, and therefore the whole conditional true.
But the one offered by the book merely says: "There exists one ball and you need to have it" ... which really doesn't capture the 'so' part.  Indeed, the statement given is really an expression of an argument: [This], therefore [that]. And there is really no truth-functional connective or other nice way to logically represent a 'therefore' or 'so'.
Of course, what you can do is to represent this argument using two statements in predicate logic, and use the mathematical 'therefore' in between:
$\exists x (Ball(x) \land \forall y (Ball(y) \to x = y))$
$\therefore$ $\exists x (Ball(x) \land Have(you,x))$
A: "$\phi \to \psi$" means "not $\phi$ or $\psi$. So your formula permits for the possibility where there is not exactly one (i.e. no or more than one) ball. But we need to have that there is one ball, and that you have it. Though there is more to the semantics of "so" than just that both sides are the case; it additionally carries the meaning of a causal connection, and this additional meaning can not be expressed solely by truth-functional operators.
