Question

Prove that (A U B)' = A ∩ B'
My proof

Let x ∈ (A U B)'
So x ∈/ A or B
x∈/A→x∈ A' and x∈/B→x∈B'
Therefore
x∈ A' ∩ B'

Answer 1

A'(INTERSECTION)B'

Answer 2

right

Answer 3

ITS A IDENTITY A UNION B COMPLEMENT IS EQUAL TO A COMPLEMENT INTERSECTION B COMLEMENT

Answer 4

what does that mean?

Answer 5

\[A \cup B =A'intersection B'\]

Answer 6

\[(A \cup B)'=A' (INTERSECTION) B'\]

Answer 7

GOT OR NOT???

Answer 8

Not

Answer 9

WHAT DO YOU WANT ME TO SOLVE?

Answer 10

nothing, I was asked to prove the question I asked.

Answer 11

Have I shown it or not?

Answer 12

your proof is wrong
in second step

Answer 13

\[x \in(A \cup B)'\]

Answer 14

\[x \notin (A \cup B)\]

Answer 15

\[(x \notin A ) and ( x \notin B )\]

Answer 16

x\[(x \in A') and (x \in B')\]

Answer 17

\[x \in A' (INTERSECTION) B'\]

Answer 18

\[(A \cup B)'\subseteq A'(intersection) B'\]

Answer 19

You say
Let x ∈ (A U B)'
x ∈/ (A U B) <--- add this line. we got rid of the not operator.
So x ∈/ A or B <--- I would say x ∈/ A and x ∈/ B (if x is not in the union of A and B, x is not in A and it is not in B
x∈/A→x∈ A' and x∈/B→x∈B'
Therefore, because x is a member of both A' and B', x must be a member of their intersection
x∈ A' ∩ B'