# 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!)

• Who is Ann? :-) – psmears Nov 23 '17 at 14:57

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 $y$: $$\exists x \exists y \left[Sx \land Sy \land Lax \land Lay \land x \neq y \land ∀z(Sz \land Laz \to z=x \lor z=y) \land (Ox \lor Oy) \right]$$
• Also, we may as well require $x$ to be old, so we can simplify the disjunction to just $Ox$. – 6005 Nov 23 '17 at 10:10
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.
$\exists x\exists y (Sx \wedge Sy \wedge Lax \wedge Lay \wedge x\neq y \wedge \neg \exists z(z\neq x \wedge z\neq y \wedge Sz \wedge Laz) \wedge (Ox \vee Oy))$