Prove: $A\cap B=\emptyset \implies A\cap B^c=A$[EDITED]

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Problem: $A\cap B=\emptyset \implies A\cap B^c=A$

I have partly done the proof, but I have not idea if I am headed in the right direction. Any help would be appreciated.

Since I know that A∩B is disjoint, A∩$B^c$ is not disjoint so that x∈A & x∈$B^c$

so A⊆A∩$B^c$ & A∩$B^c$⊆A because A∩B' means that A includes all elements in A∩B' and vice versa

Where do I go from this?

Edit: I fixed the title, and found the appropriate symbols. I hope this is enough to receive help, thanks.

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If you've proved that $A \subseteq A \cap B^c$ and $A \cap B^c\subseteq A$ then you're done.

Two sets $C,D$ are equal if they have the same elements: $x \in C \iff x \in D$. This is the same as $C \subseteq D$ and $D \subseteq C$