Question

how do you prove this equation with law of theory?

Answer 1

bring in demorgans laws a couple times

Answer 2

okay that was wrong sorry

Answer 3

you can pretend im not watching XD

Answer 4

\[\lnot [( A\cap \lnot B)\cup (\lnot B\cup C)]\]
use demorgan to get
\[\lnot(A \cap \lnot B)\cap \lnot(\lnot B\cup C)\]

Answer 5

then use it again for each piece

Answer 6

get
\[(\lnot A\cap B)\cap (B \cup \lnot C)\]

Answer 7

you mean this ? ¬[(¬A∩B)∪(B∪¬C)∩(¬A∩C)]

Answer 8

then \[(\lnot A\cap B)\cap (B \cup \lnot C)\cup (A\cap \lnot C)\] is what i get

Answer 9

then rewrite the left two parentheses

Answer 10

iSee, but how do you get that step down to (B∩¬C)?
thats where it confuses me

Answer 11

\(X\cap (Y\cup Z)=(X\cap Y)\cup (X\cap Z)\)

Answer 12

so get rid of the intersection between the first two parentheses, by writing it as a union of intersections

Answer 13

\[(\lnot A\cap B)\cap (B \cup \lnot C)\cup (A\cap \lnot C)\]
\[(\lnot A\cap B\cap B)\cup (\lnot A\cap B \cap C)\cup (A\cap \lnot C)\]

Answer 14

damn a typo

Answer 15

\[(\lnot A\cap B\cap B)\cup (\lnot A\cap B \cap\lnot C)\cup (A\cap \lnot C)\]

Answer 16

now that everything is a union, you can remove the parentheses
also of course \(B\cap B=B\)

Answer 17

do you mean i can remove all the parentheses and recombine every term?

Answer 18

you can reorder them

Answer 19

so when is this reordering thing valid ? can i just do it in the very first step ?

Answer 20

no because you have the \(\lnot\) if front of the first parentheses

Answer 21

now you have nothing but unions

Answer 22

\[(\lnot A\cap B\cap B)\cup (\lnot A\cap B \cap\lnot C)\cup (A\cap \lnot C)\]
\[(\lnot A\cap B)\cup (\lnot A\cap B\cap \lnot C)\cup (A\cap \lnot C)\]
\[(\lnot A\cap B\cap \lnot C)\cup (A\cap \lnot C)\]

Answer 23

can i see ∩ and ∪ as regular operator + and -?

Answer 24

not really

Answer 25

i can see how you get down to last step, it just doesnt quite make sense too me . sigh

Answer 26

then \[(\lnot A\cap B\cap \lnot C)\cup (A\cap \lnot C)\]
\[(B\cap \lnot C)\] as
\[(\lnot X\cap Y)\cup (X\cap Y)=Y\]

Answer 27

oh maybe i made a mistake
i didn't write it out on paper first
let me do that

Answer 28

i did make a mistake, there are several more steps

Answer 29

i think we can do this, but it is going to take a minute or two, especially to write it out
it is a lot of changing stuff

Answer 30

lol, omg im feel bad for this. but thanks for doing this

Answer 31

ok first the original question
\[\lnot [( A\cap \lnot B)\cup (\lnot B\cup C)]\cup (A\cap \lnot C)\]

Answer 32

we want to write everything as a union, so we can commute
the last part we do not need to touch, because it is already \(\cup (A\cap \lnot C)\) so we leave it alone and concentrate on
\[\lnot [( A\cap \lnot B)\cup (\lnot B\cup C)]\]

Answer 33

bring the \(\lnot \) inside the parentheses by turning the \(\cap\) into a \(\cup\)

Answer 34

damn another typo

Answer 35

\[\lnot [( A\cap \lnot B)\cup (\lnot B\cap C)]\]\[=[\lnot ( A\cap \lnot B)\cap (\lnot B\cap C)]\]

Answer 36

there i finally got it right

Answer 37

now bring the \(\lnot\) inside both parentheses

Answer 38

damn i cannot seem to write this correctly
sorry

Answer 39

lol, its all right

Answer 40

\[\lnot [( A\cap \lnot B)\cup (\lnot B\cap C)]\]

Answer 41

there that is what we concentrate on
i had it wrong
okay

Answer 42

\[\lnot [( A\cap \lnot B)\cup (\lnot B\cap C)]\]
\[=[\lnot ( A\cap \lnot B)\cap \lnot(\lnot B\cap C)]\]

Answer 43

now bring in the \(\lnot\) again

Answer 44

\[=[(\lnot A\cup B)\cap (B\cup \lnot C)]\]

Answer 45

now demorgan distributive law twice

Answer 46

\[((\lnot A\cup B)\cap B)\cup (\lnot (A\cup B)\cap \lnot C)\] for the first one

Answer 47

now distribute again

Answer 48

\[(\lnot A\cap B)\cup (B\cap B)\cup (\lnot A\cap \lnot C)\cup (B\cap\lnot C)\]

Answer 49

now we can bring back the \(\cup (A\cap \lnot C)\) and reorder

Answer 50

\[(B\cap B)=B\]

Answer 51

and
\[(A\cap \lnot C)\cup (\lnot A\cap \lnot C)=\lnot C\]

Answer 52

and \((\lnot A\cap B)\subset B\)

Answer 53

and also \((B\cap \lnot C)\subset B\) and also \(\subset \lnot C\)

Answer 54

so you are left with \(B\cup \lnot C\)

Answer 55

sorry it took so long

Answer 56

nono, my you good . thanks soo much

Answer 57

yw

Answer 58

with a bagel