Prove that $A\cap B= \emptyset$ iff $A \cap B' = A$

Question

Prove $A\cap B= \emptyset$ iff $A \cap B' = A$.

I can prove its reverse, I mean $A \cap B' = A$ iff $A \cap B = \emptyset$. I can also understand why this would be true, $A \cap B = \emptyset$ iff $A \cap B' = A$.

Answer 1

$A \cap B' \subseteq A $ always true

$$ p \; \text{and} \; q \implies p $$

Answer 2

we can easily interpret this in a Venn diagram for understanding

Answer 3

$x∈A∩B$

$⇒x∈A∧x∈B$

$⇒x∈A∩B′∧x∈B[∵A∩B′=A]$

$⇒(x∈A∧x∈B′)∧x∈B$

$⇒(x∈A∧x∉B)∧x∈B$

$⇒x∈A∧(x∉B∧x∈B)$

$⇒x∈A∧x∉∅$

$⇒x∉∅$

$∴A∩B⊆∅$

$As$ $∅$ $is$ $an$ $empty$ $set$ , $∅⊆A∩B$

$∴ A∩B=∅$

$[proved]$