Hi,

if

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

thanks in advance!

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- Oct 20th 2010, 07:44 AMsunmalusNegation
Hi,

if

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

thanks in advance! - Oct 20th 2010, 08:03 AMMoeBlee
- Oct 20th 2010, 08:20 AMsunmalus
it's what I got at first but then I thought about it and the fact that

made me think that is was wrong... So last question is that negation right or is it ??

logically I'd say

but once again it seems weird... - Oct 20th 2010, 08:28 AMMoeBlee
No, it should be:

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:

" " stands for " ".

Anyway, all you need is:

- Oct 20th 2010, 08:44 AMMoeBlee
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 AMsunmalus
Last questions:

is not the same as:

right?

If not what is another way to represent

(I think this was my real problem the whole time) - Oct 20th 2010, 10:49 AMMoeBlee
That makes no sense. It is not well formed, not syntactically correct.

Probably what you mean is this:

x((x A & x B) -> x A B)

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

(x A & x B) -> x A 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 AMsunmalus

this is what I meant but I thought it was clear because of all the previous posts, sorry. - Oct 20th 2010, 11:21 AMemakarov
It should be pointed out that is not a well-formed proposition (formula). The inductive definition of formulas goes like this. "... If P and Q are formulas, then is a formula... If P(x) is a formula that may contain a free variable x, then is a formula." So something that starts with is not a well-formed formula. (The culprit, of course, is

*not*.)

If you write the statement as

reasoning about it would be easier. - Oct 20th 2010, 11:24 AMMoeBlee
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 AMsunmalus
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 PMMoeBlee
delete post