Why implication rather than conjunction while translating universal quantifiers?

Question

Consider the following statements:

R(x): x has taken a course in Rhetoric
S(x): x is a student in this class

"Some student in this class has taken a course in Rhetoric" is translated as:
∃x (S(x) ∧ R(x))

okay I get that.

However, "Every student in this class has taken a course in Rhetoric" is translated as:
∀x (S(x) → R(x))
and not
∀x (S(x) ∧ R(x))

I am having a tough time understanding why ∀x (S(x) ∧ R(x)) is the wrong representation. Apparently it means something else but for me intuitively ∀x (S(x) ∧ R(x)) conveys the meaning "Every student in this class has taken a course in Rhetoric".

Kindly also provide an intuitive explanation why the symbolic representation for the two quantifiers is not similar (i.e. one uses conjunction and the other uses implication).

Answer 1

This is a typical question when one tries to formalize natural language statements with logic propositions and connectives. The key is that the antecedent in any material conditional is only hypothetical (if...then...) which is exactly what's needed in a universal quantification typically. Otherwise usually it's not what you mean originally.

For example, let x ranges over all things. Let R(x) mean x is a human and S(x) x will die. If you want to translate "All humans will die." ∀x (S(x) ∧ R(x)) is incorrect, because now it reads as "Anything is human and will die." Only the conditional proposition preserves the original English meaning.

Same logic for your student taking class example. The S(x) in your problem only means "x is a student in this class", but the domain of discourse is all students some of whom may not be in this class at all.

Finally if your domain of discourse is just all students in this class, then your correct formal sentence should be written only as ∀x R(x), you'll intuitively understand in such case there's no need of conjunction or connective at all. Conclusion: fully aware of your domain of discourse matters. At the root this is not a math problem, it's an epistemic one. It's like you first need to clarity and understand all definitions and context of a problem before solving the problem correctly and intelligently. Philosophy matters before math.

Answer 2

We wish to say: "Everyone takes the course if they are a student-in-the-class". That is the implication $\forall x~(S(x)\to R(x))$

We do not wish to claim: "Everyone is a student-in-the-class and they take the course". That is the conjunction $\forall x~(S(x)\land R(x))$.

After all, there may be someone who is not a student-in-the-class. We wish to allow this, and to not make any claim about them.

So we go with the implication: $\forall x~(S(x)\to R(x))$

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For the existential, the claim is "there is a student-in-the-class and they take this course." Thus it uses the conjunction: $\exists x~(S(x)\land R(x))$

Answer 3

It's important to realize that first-order logic has a single domain of discourse. This means that every entity that exists is in the domain that quantifiers can range over.

With this in mind, the well-formed formula (110) means (111) and does not mean (130) or (131).

$$ \forall x \mathop. S(x) \land R(x) \tag{110} $$

$$ \text{Every being is a student in this class and has taken rhetoric}. \tag{111} $$

(130) and (131) are not equivalent to (110). However, (130) is equivalent to (131).

$$ \text{For every being $x$, if $x$ is a student in this class, then $x$ has taken} \tag{130} $$ $$ \text{Every student in this class has taken rhetoric} \tag{131} $$

With this in mind, we can see that (130) and (131) can be expressed as (132) or (133).

$$ \forall x \mathop. S(x) \to R(x) \tag{132} $$ $$ \forall x \mathop. (\lnot S(x)) \lor R(x) \tag{133} $$

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I think the reason that your intuition is misleading you is that the sentence "Every student in this class has taken a course in Rhetoric" is talking about a collection of beings that are all students. In an intuitive sense, it feels like their studentness and in-this-class-ness is being asserted.

One way of capturing this intuition is to use the notion of a bounded quantifier, for example in (134) below.

$$ \forall x \in S \mathop. R(x) \tag{134} $$

If we expand the domain from $S$ to the set of all entities, then we need to make sure that non-students do not affect the truth value of the statement.

In order for non-students to not affect the truth value of the statement, we need to make sure that the body of the quantifier expression has the value true when $x$ is a non-student. We also need to make sure that the body of the quantifier expression has the value false when a student in this class has not taken rhetoric. This is how we get the translation of the statement.

$$ \forall x \mathop. (\lnot S(x)) \lor R(x) $$

Answer 4

The error is even worse than it might seem.

If you just write $\forall x$ followed by some formula with variable $x,$ the $x$ could be anyone or anything. A student in your school is an $x,$ but also I am an $x,$ Joe Biden is an $x,$ some kid's pet hamster is an $x,$ and the spare tire in my car trunk is an $x.$

When you write $\forall x.(S(x)\land R(x))$ you are asserting (among other things) that my spare tire is a student in your class and has taken a course on rhetoric.

Be careful with universal quantifiers.

Answer 5

• “Every Student in this class has taken a course in Rhetoric.” (left)

  All members of $S$ also belong to $R.$

versus

• “Every person is a Student in this class who has taken a course in Rhetoric.” (right)

  All members of the universe lie in $R\cap S.$

Legend:
set $P$ contains variable $x$ $\quad:=\quad$ predicate $P(x)$ is true
set $P$ is shaded $\quad:=\quad$ predicate $P(x)$ is never true