Can you proof this ?

can you proof this ?

can you english?

Yes, you also should be capable except in the cases of massive brainletism.

Fucking discrete math. Just applying definitions ad nauseam.

Thanks for your help

Yes

what does set x set mean?

yes

wlog for all x they be in A and not in B
then for any (x, b), b in B, it will be that for (b, x) will be not.

now other way:
wlog for all (x, b), b in B, they be in (A x B) but not in (B x A).
so in (B x A) there can be no x as first term for all x.
So for all x they be in A and not in B.

i can not very good english

I proved this in like 30 seconds in my mind and so should you. You shouldn't be asking for help on such trivial stuff.

I've never taken discrete maths but it looks like it can be done using Demorgan's Laws?

I understand truth behind the statement... I just dont know how to write it down

Then you don't understand it.

why are they writing =!0 when they could just write =1?

> he hasn't read Boole's notation on temporality logic user
kek

forgot pic.

Ok ok.

First, suppose [math] A \cap B = \varnothing [/math] .
Aiming for contradition, suppose that these products had a common element. That would mean this common element is [math] (a,b)=(b',a') [/math] , which means that [math] A \ni a=b' \in B [/math] which means that [math] A \cap B \neq \varnothing [/math]

Now we prove the other direction:
Suppose that the intersection of the products is empty.
Aiming for contradiction, suppose that A and B have a common element x.
Then (x,x) is in both products, making their intersection non-empty.

this helped, thank you

np

cartesian product

Set of tuples where n-th element is from n-th set. So AxB is a set of 2-tuples (a,b) where a is from A and b is from B .

(=>) c in A&B => (c,c) in (AxB)&(BxA)

( c in A & c in B => c in A&B

out CS major.

not really, it just means the communication units are inadequate or unknown.

If he can't prove something is true, he simply doesn't understand why it is true. Mistaking intuition for understanding is dangerous.