Existential quantifier I've been doing a bit of independent research on existential quantification, and I was wondering if anyone could tell me if I have the hang of it. Here's a statement regarding the intermediate value theorem of calculus:

If $f$ is a function of $x$ that is continuous on the interval $[a,b]$, then $$\exists\left( p\in [a,b]\right)\,\left( f(p)\in [f(a),f(b)]\right)$$ is true.

If it helps, I typed this as
If $f$ is a function of $x$ that is continuous on the interval $[a,b]$, then $$\exists\left( p\in [a,b]\right)\,\left( f(p)\in [f(a),f(b)]\right)$$ is true.
If you are having trouble with existential quantifiers, there are just four rules that govern its usage in a mathematical proof (with each quantifier explicitly restricted to a set):


From $\exists x\in A: P(x)$, we can infer that $y\in A$ and $P(y)$ where $y$ is a free variable not previously used in that proof. (Existential specification or instantiation.)
(EDIT) From $y\in A$ and $P(y)$, we can infer that $\exists y\in A: P(y)$, or that $\exists x\in A: P(x)$ where $x$ is not a bound variable in $P(y)$, and where $P(x)$ is formed replacing some or all occurrences of $y$ in $P(y)$ with $x$. (Existential generalization)
From $\exists x\in A: P(x)$, we can infer that $\neg \forall x\in A:\neg P(x)$. (Changing quantifier, $\exists \to \forall$)
From $\forall x\in A:P(x)$, we can infer that $\neg\exists x\in A: \neg P(x)$. (Changing quantifier, $\forall \to \exists$)


where $A$ is a set and $P$ is a unary predicate.

Note that: $$\exists x\in A: P(x) \equiv \exists x: [x\in A \land P(x)]$$  $$\forall x\in A: P(x) \equiv \forall x: [x\in A \implies P(x)]$$

Avoid the construct:

$$\exists x: [x\in A \implies P(x)]$$ 

Very strange things can happen otherwise! (See "Drinker's Paradox" at Wikipedia and at my blog.)
If you are having trouble with existential quantifiers, there are just four rules that govern its usage in a mathematical proof (with each quantifier explicitly restricted to a set):


From $\exists x\in A: P(x)$, we can infer that $y\in A$ and $P(y)$ where $y$ is a free variable not previously used in that proof. (Existential specification or instantiation.)
(EDIT) From $y\in A$ and $P(y)$, we can infer that $\exists y\in A: P(y)$, or that $\exists x\in A: P(x)$ where $x$ is not a bound variable in $P(y)$, and where $P(x)$ is formed replacing some or all occurrences of $y$ in $P(y)$ with $x$. (Existential generalization)
From $\exists x\in A: P(x)$, we can infer that $\neg \forall x\in A:\neg P(x)$. (Changing quantifier, $\exists \to \forall$)
From $\forall x\in A:P(x)$, we can infer that $\neg\exists x\in A: \neg P(x)$. (Changing quantifier, $\forall \to \exists$)


where $A$ is a set and $P$ is a unary predicate.

Note that: $$\exists x\in A: P(x) \equiv \exists x: [x\in A \land P(x)]$$  $$\forall x\in A: P(x) \equiv \forall x: [x\in A \implies P(x)]$$

Avoid the construct:

$$\exists x: [x\in A \implies P(x)]$$ 

Very strange things can happen otherwise! (See "Drinker's Paradox" at Wikipedia and at my blog.)