Express each of the following statements as expression using quantified predicates and the domain"People."

Question

Here are two questions confused me.

Express each of the following statements as expression using quantified predicates and the domain "People."

1) Some high school students are not enrolled in class Calculus II.

My Try:

$S(x)$ means $x$ is a high school student, for $x \in \text{People}$

$E(x,y)$ means $x$ is enrolled in class $y$, for $x\in \text{People}$, and $y\in \text{Classes}$

$$\exists x ~:~ (S(x) \land \lnot E(x, \text{Calculus II})), x\in \text{People}$$

What I am not sure here is can I put the class title into the quantified predicates directly?

2) Every primary school student is younger than any high school student.

My Try:

$P(x)$ means $x$ is a primary school student, for $x\in \text{People}$

$H(x)$ means $x$ is a high school student, for $x\in \text{People}$

$Y(x,y)$ means $x$ is younger than $y$, for $x,y \in \text{People}$

So I got: $\forall x~:~((P(x) \land \forall y~:~(H(y))) \implies Y(x,y)) ~.~ x,y\in \text{People}$

before this solution, I tried

$$\forall x ~:~ \forall y ~:~ (P(x) \land H(y) \land Y(x,y)) ~.~ x,y\in \text{People}$$

But someone else notify me that using implication is more make sense. I am so confusing about why. And should I put $\forall y~:~H(y)$ inside?

Answer 1

For Part (1), I suggest instead of using $E(x, y)$, just use $C(x)$ and have it mean that a student $x$ is in Calc II. But your answer is correct still.

For part 2, writing

$$\forall x ~:~((P(x) \land \forall y ~:~(H(y))) \implies Y(x,y))$$

Means "Every person is a primary school student, and every high school student is older than him". That is not what you intend.

$$\forall x ~:~ \forall y ~:~ (P(x) \land H(y) \land Y(x,y))$$

This means "Every person is a primary student, and every person is a high school student, and every pair of people is younger than each other". That is also probably not what you intend.

Let's try "Every pair of people, if the first is a primary student and the second is a high school student, then the first is younger than the second.":

$$\forall x ~:~ \forall y ~:~ (P(x) \land H(y)) \implies Y(x, y)$$

which is:

$$\begin{array} {c} \forall x ~:~ \forall y ~:~ & \text{For every pair of people} \\ (P(x) & \text{The first is a primary student} \\ \land & \text{and} \\ H(y)) & \text{the second is a high school student} \\ \implies & \text{(If) then} \\ Y(x, y) & \text{The first is younger than the second} \end{array}$$

Note that your question specifies that the domain is people. So $\forall x ~:~ \forall y ~:~$ means for every pair of people $(x, y)$. It is incorrect to put "$, x,y \in \text{People}$" at the end of your statement: for one, "$.$" and "$,$" are not logical operators. It is a grammatical mistake to use them. Also, it is redundant, it is like saying "for every pair of integers that is a pair of integers...". You can just acknowledge that the domain is people and leave it out of formulas.

Answer 2

$\exists x ~:~ (S(x) \land \lnot E(x, \text{Calculus II})), x\in \text{People}$

Two notes:

1. While defining the predicates, you may as well use $C$ for "is enrolled in Calculus II"; particularly since the domain of discourse is "people" then such should be the only variables and constants in play.
2. The domain restriction should be attached directly to the quantifier, or be left implied (but stated at the start of the discussion).   If explicit, assign it a token along with the predicates: Let $\mathsf P$ be the domain of "People".

$$\exists x\in \mathsf P:\big(S(x)\wedge \neg C(x)\big) \\ \exists x:\big(S(x)\wedge \neg C(x)\big)$$

$\forall x~:~((P(x) \land \forall y~:~(H(y))) \implies Y(x,y)) ~.~ x,y\in \text{People}$

... No.   The implication is the right track, but the quantifier of $y$ does not belong inside the antecedent of the implication.   Among other reasons, the predicate $Y$ is then outside of that quantifier's scope.

To properly scope all instances of the $y$ variable, lift it into consequent.

$$\forall x{\in}\mathsf P :\Big(P(x)\to \forall y{\in}\mathsf P:\big(H(y)\to Y(x,y)\big)\Big)$$

Alternatively the prenex form is:

$$\forall x{\in}\mathsf P~\forall y{\in}\mathsf P:\Big(\big(P(x)\wedge H(y)\big)\to Y(x,y)\Big)$$