Logical meaning of for all Quantifier

Question

Let S(x) be the statement "x is a student in the class".

C(x) be "x has studied Calculus".

The domain consist of all the people.

The statement "Every student in the class has studied calculus" is expressed as

∀x(S(x) -> C(x)). (in the book K.H. Rosen)

Now for the case where S(x) is False and C(x) is true, the expression is still true (according to the truth table of "->") which states the statement: "All people who're not students of the class have studied calculus". How come this statement is equivalent to the above statement "Every student in the class has studied calculus"

Answer 1

You are misinterpreting the sentence: ∀x(S(x) -> C(x)). You can read it as "For every x, if x is a student, then x have studied calculus".

It doesn't make sense to ask if S(x) is False or not since x, being in the reach of the "for all" cuantifier, ranges over the Universal set (the set of people in this case).

To probe that the sentence ∀x(S(x) -> C(x)) is True you need to probe that the sentence

                               S(x) -> C(x)

is True for every x, ie for each person in the set of people.

You should try to write down the statment "All people who're not students of the class have studied calculus", or an easier equivalnet "For all people, if they are not students, then they have studied calculus" and compare with the original.

Answer 2

If for some $x$, it is the case that $S(x)$ is false and $C(x)$ is true, then indeed $S(x)\to C(x)$ is true, this tells us that for ths $x$ also $(\neg S(x))\to C(x)$ is true. But a single example does not warrant $\forall x((\neg S(x))\to C(x))$ (which is the formalization of "All people who're not students of the class have studied calculus").

Answer 3

We have different scenarios in place...

Consider for simplicity the domain $D= \{ John, Rachael, Cody \}$.

Scenario 1) :

$S$ = students in the class = $\{ John, Rachael, Cody \}$

and

$C$ = students of calculus = $\{ John, Rachael, Cody \}$.

We have that $S(John) \to C(John)$ is $T \to T$, i.e. $T$, and this holds for all three.

Thus, in Scenario 1), we have that :

$∀x(S(x) \to C(x))$ is True.

Scenario 2) : $S = \{ John, Rachael, Cody \}$ and $C = \{ Rachael, Cody \}$.

We have that $S(John) \to C(John)$ is $T \to F$, i.e. $F$, and thus :

$∀x(S(x) \to C(x))$ is False.

Finally : Scenario 3) : $S = \{ Rachael, Cody \}$ and $C = \{ Rachael, Cody \}$.

We have that $S(John) \to C(John)$ is $F \to F$, i.e. $T$. And we have $T$ also for $Rachel$ and $Cody$.

Thus, in Scenario 3), we have again that :

$∀x(S(x) \to C(x))$ is True.

Answer 4

Now for the case where S(x) is False and C(x) is true, the expression is still true (according to the truth table of "->") which states the statement: "All people who're not students of the class have studied calculus". How come this statement is equivalent to the above statement "Every student in the class has studied calculus"

As your truth table shows, $\forall x~(S(x)\to C(x))$ is equivalent to $\forall x~(\neg S(x)\lor C(x))$ and also $\forall x ~\neg(S(x)\wedge\neg C(x))$.

That says, an aritrary person will (a) be not a student in the class $\underline{\textit {or}}$ (b) have studied calculus.   That is there will not be any students in the class who has not studied calculus.

Clearly that means everyone who is a student in the class must have studied calculus.