$(\forall x\in S)P(x)\equiv \forall x(x\in S\Rightarrow P(x))$ What does it mean?

Question

$(\forall x\in S)P(x)\equiv \forall x(x\in S\Rightarrow P(x))\\(\exists x\in S)P(x)\equiv \exists x(x\in S\wedge P(x))$

What I think it means:

1) For all $x$ in $S$, $P(x)$ holds true = for all $x$, such that, if $x$ is in $S$, then $P(x)$ holds true. Does this mean that $(x\in S\Rightarrow P(x))$ is always true, therefore, if $x$ is indeed in $S$, $P(x)$ has to be true, for the expression to evaluate to true? (Because $T \Rightarrow T \equiv T$.)

I think I'm wrong, so if someone could clarify this...

Also, what would change if instead of 'implies' operator we had 'and' operator (like in the second line)? It again had to be true so the expression evaluated true.

Answer 1

$$(\forall x\in S)P(x)\equiv \forall x(x\in S\Rightarrow P(x))$$

First note that the connective $\equiv$ means that if one side of the connective is true, so is the other (and if one side is false, so is other). Put differently, it means that $$(\forall x\in S)P(x)\text{ if and only if} \forall x(x\in S\Rightarrow P(x))$$

So yes, both sides declare that for each $x$ if $x\in S$, then $P(x)$.

The implication in the right hand side cannot be changed to $\land$, because if it were, the equivalence would fail.

Note the difference:

For all numbers $x$, if $x$ is an integer, then $x+1$ is an integer.

The consequent applies only to integers, and if $x$ is, say, a fraction (like 1/2), then statement is vacuously true, since the antecedent is false, hence the implication is always true.

Versus

For all numbers $x$, $x$ is an integer and $x+1$ is an integer. This is blatantly false for all $x \notin \mathbb Z$.

That is, suppose we have the statement $\forall x(x\in \mathbb Z \land x+1\in \mathbb Z)$.

The statement is true only if $x \in \mathbb Z$. But the statement above says that for every (number, object, triangle, etc): It is an integer AND it plus the number 1 is also an integer. Clearly, this is absurd.

Answer 2

Yes, the expression $(∀x∈S)P(x)$ is the so-called restricted or relativized (universal) quantifier, and it is precisely equivalent to :

$\forall x(x \in S \Rightarrow P(x))$.

If you consider that we can use a predicate $S(x)$ such that $S(x)$ holds iff $x \in S$, then the last formula is equivalent to a "pure" predicate formula :

$\forall x(S(x) \Rightarrow P(x))$.

Now, it must be clear that the usual "rules" for evaluating quantified formulae apply.

Thus, the formula "for all $x$, if $x$ is in $S$, then $P(x)$" is true when all $x$ in $S$ are also $P$.

This formula is the "logical form" of the traditional "all men are mortal", which states that the class of men is included into the class of "mortals".

The same reasoning applies to the existential one :

$(∃x∈S)P(x)$.

It can be re-written as :

$∃x(S(x) \land P(x))$

and it is true precisely when there is some $x$ such that is both $S$ and $P$.

Note

Why $\forall x(x \in S \Rightarrow P(x))$ has not the same meaning of $\forall x(x \in S \land P(x))$ ?

Consider again $\forall x(S(x) \Rightarrow P(x))$, which means "all $S$ are $P$".

Its negation will be : "not all $S$ are $P$", which we can equivalently express as : "there are some $S$ which are not $P$" [try with : "not all men are mortals" is the same as : "there are some man which is immortal (i.e. non-mortal)"].

Consider now the negation of $\forall x(S(x) \Rightarrow P(x))$, i.e.

$\lnot \forall x(S(x) \Rightarrow P(x))$

which is (by rules for quantifiers) :

$\exists x \lnot (S(x) \Rightarrow P(x))$.

But $p \Rightarrow q$ is equivalent to : $\lnot p \lor q$; thus, applying this "transformation to the above formula and using De Morgan, we get :

$\exists x (S(x) \land \lnot P(x))$

which translates well "there are some $S$ which are not $P$". In particular, the formula is false if there are no $S$.

Consider now the purported "alternative" translation of $(∀x∈S)P(x)$ as $\forall x(x \in S \land P(x))$.

If we consider the negation of the last formula we get :

$\exists x \lnot (S(x) \land P(x))$

which is (again by De Morgan) :

$\exists x (\lnot S(x) \lor \lnot P(x))$.

This is not equivalent to the previous translation, because this formula is true if there are no $S$.

Conclusion : if the contradictory of the two formulae have not the same "meaning" the two formulae are not equivalent.