# Negation

• Oct 20th 2010, 07:44 AM
sunmalus
Negation
Hi,

if $\displaystyle y \in f( \mathbf{A} \cup \mathbf{B}) \Leftrightarrow \exists x \in \mathbf{A} \text{ or } \exists x \in \mathbf{B} \text{ such as }y=f(x)$

$\displaystyle \neg(y \in f( \mathbf{A} \cup \mathbf{B}))= y \not \in f( \mathbf{A} \cup \mathbf{B}))\Leftrightarrow \forall x \not \in \mathbf{A} \text{ and } \forall x \not \in \mathbf{B} \text{ we have }y=f(x)$

it seems right to me but the last part seems weird "we have"...

• Oct 20th 2010, 08:03 AM
MoeBlee
Quote:

Originally Posted by sunmalus
$\displaystyle \forall x \not \in \mathbf{A} \text{ and } \forall x \not \in \mathbf{B} \text{ we have }y=f(x)$

That's not right. It should be:

$\displaystyle \forall x \in \mathbf{A} \text{ and } \forall x \in \mathbf{B} \text{ it is not the case that }y=f(x)$
• Oct 20th 2010, 08:20 AM
sunmalus
it's what I got at first but then I thought about it and the fact that
$\displaystyle \neg(y\in f(\mathbf{A} \cup \mathbf{B}))=y \not \in f(\mathbf{A} \cup \mathbf{B})$
made me think that is was wrong... So last question is that negation right or is it $\displaystyle \neg(y\in f(\mathbf{A} \cup \mathbf{B})=y \not \in f(\mathbf{A} \cap \mathbf{B})$??

logically I'd say
$\displaystyle \neg(y\in f(\mathbf{A} \cup \mathbf{B}))=y \not \in f(\mathbf{A} \cup \mathbf{B})$
but once again it seems weird...
• Oct 20th 2010, 08:28 AM
MoeBlee
Quote:

Originally Posted by sunmalus
$\displaystyle \neg(y\in f(\mathbf{A} \cup \mathbf{B}))=y \not \in f(\mathbf{A} \cup \mathbf{B})$

No, it should be:

$\displaystyle \neg(y\in f(\mathbf{A} \cup \mathbf{B})) \Leftrightarrow y \not \in f(\mathbf{A} \cup \mathbf{B})$

In this kind of context, an equal sign doesn't go between two statements. Rather, to express that two statements are equivalent, use the biconditional sign.

Or (and this is a picky technical point having not much substantive to do with the subject of images that these formulas are about), you could say:

"$\displaystyle y \not \in f(\mathbf{A} \cup \mathbf{B})$" stands for "$\displaystyle \neg$ $\displaystyle y\in f(\mathbf{A} \cup \mathbf{B})$".

Anyway, all you need is:

$\displaystyle \neg$ $\displaystyle y\in f(\mathbf{A} \cup \mathbf{B}) \Leftrightarrow y \not \in f(\mathbf{A} \cup \mathbf{B})$
• Oct 20th 2010, 08:44 AM
MoeBlee
Darn, I needed to edit my previous post because I messed up MATH tags. Please read post again now edited.

Anyway, without messing with tags, in plain ASCII:

~ y in f[A u B] <-> y not in f[A u B]
• Oct 20th 2010, 10:30 AM
sunmalus
Last questions:
$\displaystyle \forall x \in \mathbf{A} \cap \forall x \in \mathbf{B}$

is not the same as:

$\displaystyle \forall x \in \mathbf{A} \cap \mathbf{B}$

right?

If not what is another way to represent
$\displaystyle \forall x \in \mathbf{A} \cap \forall x \in \mathbf{B}$

(I think this was my real problem the whole time)
• Oct 20th 2010, 10:49 AM
MoeBlee
Quote:

Originally Posted by sunmalus
$\displaystyle \forall x \in \mathbf{A} \cap \forall x \in \mathbf{B}$

That makes no sense. It is not well formed, not syntactically correct.

Probably what you mean is this:

$\displaystyle \forall$x((x$\displaystyle \in$A & x$\displaystyle \in$B) -> x $\displaystyle \in$A$\displaystyle \cap$B)

Or even just leave off the initial universal quantifier and let it be understood as tacit:

(x$\displaystyle \in$A & x$\displaystyle \in$B) -> x $\displaystyle \in$A$\displaystyle \cap$B

I think you need to review basic symbolic logic to see how formulas are formed. I would suggest 'Logic: Techniques Of Formal Reasoning' by Kalish, Montague, and Mar.
• Oct 20th 2010, 10:53 AM
sunmalus
$\displaystyle \neg y \in f( \mathbf{A} \cup \mathbf{B})\underbrace{=}_{\text{this should be} \Leftrightarrow} y \not \in f( \mathbf{A} \cup \mathbf{B}))\Leftrightarrow \underbrace{\forall x \in \mathbf{A} \text{ and } \forall x \in \mathbf{B}}_{\not \Leftrightarrow \forall x \in \mathbf{A} \cap \mathbf{B}} \text{ is not the case that }y = f(x)$

this is what I meant but I thought it was clear because of all the previous posts, sorry.
• Oct 20th 2010, 11:21 AM
emakarov
It should be pointed out that $\displaystyle \forall x \in \mathbf{A} \text{ and } \forall x \in \mathbf{B} \text{ it is not the case that }y=f(x)$ is not a well-formed proposition (formula). The inductive definition of formulas goes like this. "... If P and Q are formulas, then $\displaystyle P\land Q$ is a formula... If P(x) is a formula that may contain a free variable x, then $\displaystyle \forall x.\,P(x)$ is a formula." So something that starts with $\displaystyle \forall x\in A\land{}$ is not a well-formed formula. (The culprit, of course, is not $\displaystyle \in A$.)

If you write the statement as

$\displaystyle (\forall x\in A.\,y\ne f(x))\land(\forall x\in B.\,y\ne f(x))$

reasoning about it would be easier.
• Oct 20th 2010, 11:24 AM
MoeBlee
No, you repeated a previous error.

(I'm tired of messing with LaTex tags, so I'm going to give you what you want in plain ASCII

~ stands for negation
e stands for membership
<-> will stands for material equivalence
u stands for binary union
U stands the universal quantification
v stands for disjunction
-> stands for material implication
not= stands for inequality)

~ y e f[AuB] <-> y not in f[AuB] <-> Ux((x e A v x e B) -> ~ y=f(x)) <-> UxeA UxeB y not= f(x) <-> Ux(x in AuB -> ~ y=f(x))

In other words:

it is not the case that y is in f[AuB]

if and only if

y is not in f[AuB]

if and only if

for all x, if (x in A or x in B) then ~ y=f(x)

if and only if

for all x in A and all x in B, ~ y=f(x)

if and only if

for all x in AuB, ~ y=f(x)

Take my suggestion to brush up on your symbolic logic?
• Oct 20th 2010, 11:31 AM
sunmalus
Thank to you all but special thanks to MoeBlee! I really get it now and yes I'll take your suggestion :).
• Oct 20th 2010, 12:10 PM
MoeBlee
delete post