# Logic, negation of a statement containing quantifiers

1. Dec 20, 2012

### SithsNGiggles

Hi, I've got another answer I'd like checked. I'm pretty sure it works out, but I want to be certain.

1. The problem statement, all variables and given/known data
Write a sentence in everyday English that properly communicates the negation of each statement.

"Some differentiable functions are bounded."

2. Relevant equations

3. The attempt at a solution
First, I wrote the statement symbolically:
$(\exists f(x) \in X) (f(x) \; \mbox{is bounded} \wedge f(x) \; \mbox{is differentiable})$,
where I let $X$ be the set of differentiable functions.

$\neg (\exists f(x) \in X) (f(x) \; \mbox{is bounded} \wedge f(x) \; \mbox{is differentiable})$

$(\forall f(x) \in X) \neg (f(x) \; \mbox{is bounded} \wedge f(x) \; \mbox{is differentiable})$

$(\forall f(x) \in X) (f(x) \; \mbox{is not bounded} \vee f(x) \; \mbox{is not differentiable})$

My question is, can I simplify this sentence to
$(\forall f(x) \in X) (f(x) \; \mbox{is not bounded})$,
since all $f(x) \in X$ are differentiable and therefore cannot be differentiable?

I just want to make sure my reasoning works here. Thanks!

As for the English translation:
"Every differentiable function is not bounded."

Last edited: Dec 20, 2012
2. Dec 20, 2012

### pasmith

So the condition "f(x) is differentiable" is equivalent to $f(x) \in X$ and so redundant:
$$(\exists f(x) \in X) (f(x)\mbox{ is bounded})$$

Yes: "P or false" is equivalent to P.

3. Dec 20, 2012

### HallsofIvy

If you are defining X to be the set of all differentiable functions, then there is no need for "$\and \text{f is differentiable}$" in your original statement. With that definition of X, your statement is simply "$(\exist f\in X)(f \text{is bounded})$" and it negation is "$(\all f\in X) (f \text{is not bounded})$".

In any case, the negation of "some differentiable functions are bounded", in "every day English" is "no differentiable functions are bounded".