Predicate Logic: Definite Quantification Regarding numerically definite quantification in predicate logic, I am struggling to formulate the following sentence due to it's complexity:
"Ada likes exactly two strong people, at least one of whom is old"
(Where $a$: Ada, $Lxy$: $x$ likes $y$, $Sx$: $x$ is strong, $Ox$: $x$ is old)
I have translated it to:
$\exists x\,\exists y\,(((Sx\land Sy)\land x\neq y)\land ∀z\,(Sz\to z=x\lor z=y)\land \exists x\,\exists y\,(Ox\lor Oy)\land (Lax\land Lay)$
I am fairly confident that my translation of 'There are at exactly two strong people' is correct:
$\exists x\,\exists y\,(((Sx\land Sy)\land x\neq y)\land ∀z\,(Sz\to z=x\lor z=y))$.
However I am questioning whether my sentence is ordered correctly so that it conveys the fact that one of the strong people is old but that Ada likes both of the strong people.
I would appreciate any help clarify this, thank you.
(I am beginner in logic so please don't judge my attempt too harshly!)
 A: As you mention, your translation of "There are exactly two strong people" is correct... However... The sentence which you need to translate does not state this. The sentence does only state that "out of all strong people, Ada likes exactly two." So this is the first error in your tranlsation. 
The second error is that you first state $\exists x\exists y$ and start talking about strong people as $x$ and $y$, but then you again say $\exists x\exists y$ as a new part of a conjunction. This means that $x$ and $y$ will be rebound and does not necessarily have to be strong people any more.
So if we restart we talk about the x and y state that they are old, ada likes them and then, only state that there are no other strong people which ada likes as one clause.
$\exists x\exists y (Sx \wedge Sy \wedge Lax \wedge Lay \wedge x\neq y \wedge \neg \exists z(z\neq x \wedge z\neq y \wedge Sz \wedge Laz) \wedge (Ox \vee Oy))$
A: It's nearly there - just don't add the extra quantifiers for $x$ and $y$, and make sure that if there was a third strong person who Ada likes, then the third person must be $x$ or $y$:
$$
\exists x \exists y \left[Sx \land Sy \land Lax \land Lay \land x \neq y \land ∀z(Sz \land Laz \to z=x \lor z=y) \land (Ox \lor Oy) \right]
$$
