Question

http://math.stackexchange.com/questions/654115/prove-a-subseteq-b-cap-c-if-and-only-if-a-subseteq-b-and-a-subseteq-c

Answer 1

...............

Answer 2

XD

Answer 3

At least I'm trying man. I'm reading more about it.

Answer 4

*thumbs up*

Answer 5

Hate it when stuff is due wed... and I'm just reading some stuff. I think either I have to switch the order of the subset like that's allowed for the proposition to work

Answer 6

Hmmm who can help you?

Answer 7

only people on stackexchange it seems... if more newbs come on here, dey be like omg wt h is diz

Answer 8

like me!!!!!

Answer 9

what are you trying to prove?

Answer 10

I am trying to prove a biconditional statement. since latex is kind of dead... I've used stackexchange to do the latex

Answer 11

LIke I can easily identify the propositions, defitinions, givens... but afterwards, my brain explodes.

Answer 12

yeah, it pretty much sucks
are you trying to prove the statement written above?

Answer 13

I think I need to use that set definition .

Answer 14

yes and there's another part to it as well. It's easy to identify the given parts and rules... applying it yes does suck

Answer 15

I got a strange feeling that i have to apply that set rule. somewhere

Answer 16

I don't think the truth table matters that much... it's the rules of the sets. though it's fun to draw the truth table. at least I still know what the meanings are.

Answer 17

you have to prove it both ways
1) if A is contained in B and C then A is contained in B and A is contained in C
stated that way it is more or less obvious, but you can do it element wise
2) if A is contianed B and A is contained in C then A is contained in B and C

Answer 18

but ok do I apply Definition 3.1.2 to this?

Answer 19

crud there's also another one too for the union and intsersection.. Definition 3.2.1 which states the sets of intersection and union. . . well the notation of it that is.

Answer 20

i don't know what Def 3.1.2 is, but the first way you can say
if an arbitrary x is in A, then since A is contained in B and C then x is in B and x is in C therefore A is contained in B and A is contained in C

Answer 21

def 3.1.2 is listed in the link

Answer 22

Definition 3.1.2 states that we should let A and B be sets. Then A is a subset of B,A⊆B, and the statement (∀x)[x∈A→∈B] is true.

Answer 23

like if A has the values 4 5 and be also has the values 4 5 then the (∀x)[x∈A→∈B] is true.

Answer 24

that is pretty much what i said in words right?

Answer 25

but if we have A with values 5 6 7 and B 3 then it's not true. oh yeah

Answer 26

I guess well I think the easiest way if I could draw with would be three circles with sets like A = 1 2 3 4 5 B = 4 5 6 7 C 7 8 9

Answer 27

but that's just drawing not really proving much...

Answer 28

i am lost sorry. i don't know what "A with values 5 6 7 and B 3 then it's not true" means

Answer 29

like say there is a set A with the numbers 5 6 7 and set B with just 3. There is nothing in common with those guys.

Answer 30

they can't be subsets of each other... no elements in common

Answer 31

hmm so wait if we go backwards on it since biconditional statements are two parts...



If A⊆B∩C then A⊆B and A⊆C

If A⊆B and A⊆C then A⊆B∩C.
like If A is contained in B and A is contained in C then A is contained in B intersecting C

Answer 32

@wio

Answer 33

Use the definition of subset and intersection to prove this.

Answer 34

that's the definition that I've provided in the url... I don't know how to apply it... well actually I attempted to apply it in the url ... definition of intersection .. oh that's Definition 3.2.1. part of it which states that the intersection of A and B written A intersection B is the set A intersect B = x x such that A and x such that B...

Answer 35

hmm do I just plug it in to the definitions and ...

Answer 36

Yes, that is all you have to do.

Answer 37

Also, don't got to stack exchange for tutoring. Most of the people there are pedantic, autistic, and perfectionist. It's not a bad place for simple questions, but not for tutoring.

Answer 38

wow O_O have you asked in stack exchange before?

Answer 39

I like the latex part since os's is broken.

Answer 40

are there any other sites that offer tutoring for higher level mathz.. . . . most places I know are like up to alg 2 or calc. >:/

Answer 41

I can tutor in higher level. If you want free, OS is best choice.

Answer 42

:D. I'm still trying to get the grip of the logic thing. like chapter 1 wasn't too bad. It's just that the lectures are moving all over the place and the homework lags like crazy so by the time I read and attempt...ughhhhh craziness. on top of that hw is due every lecture and since Tuesday is my long class days I usually stay up all night on Monday to get it done.

Answer 43

like I'm assigned 3.2.5 and 3.3.10 which is chapter 3 section 3 but there are notations from 3.1 so I'm like **** I have to backtrack.

Answer 44

We can do latex on scriblar if you want

Answer 45

:D yay! I'm gonna eat dinner, then attempt this again on paper before latexing

Answer 46

back gonna latex the attempts I made from scratch

Answer 47

\subseteq was supposed to be used x.x didn't realized until after I print screened

Answer 48

@wio

Answer 49

Use the fact that a -> p and q gives a -> p and a -> q

Answer 50

attaching the second part in a bit... I don't get why my letters are backwards

Answer 51

the last line reads C subset B and C subset A .. O_O

Answer 52

A subset C
B subset C
but the repeater C is not where A and B are

Answer 53

What is the question?

Answer 54

We have a bi-conditional statement.

If A∪B⊆C, then A⊆C and B⊆C.

If A⊆C and B⊆C, then A∪B⊆C.

Answer 55

http://assets.openstudy.com/updates/attachments/52e72f44e4b096fa405b34fb-usukidoll-1390899262784-screenshot2014012722.47.08.png

Answer 56

okay so you proved it?

Answer 57

umm not so sure on the second part.. I got a backwards result. :/

Answer 58

the first one a. looks better than b :/

Answer 59

IT seems like the only real step you do is distribute the -> over the /\

Answer 60

Here is one of them that I did.

Answer 61

In fact, I think this proof is bidirectional by default so you wouldn't need to prove it both ways.

Answer 62

crap I mixed wedge and vee up.... anyway it should be V all the way through.

Answer 63

on my latex I inputted \cap instead of \subseteq

Answer 64

This one is beast