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I made a Venn Diagram so I know that this is true. Now I just need some help on getting the proof right.

For all sets $A$ and B, if $B ⊆ A^c$ then $A ∩ B = ∅$

I have started the proof: Suppose $A$ and $B$ are set such that $B ⊆ A^c$ then WMST $A ∩ B = ∅.

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  • $\begingroup$ I really suggest my edit is put into place. $\endgroup$ – Lord_Farin Apr 5 '13 at 16:34
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If $B\subseteq \overline A$ then $b\in B$ implies $b\in\overline A$ and $b\not \in A$. So no element of $B$ is also an element of $A$ and we have $B\cap A=\varnothing$.

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Look up the definition: $A,B\subset M$ $$A^C = \{ x\in M : \ x\not \in A\}$$ For the intersection not to be empty you need that they have at least one common element.

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One way to do proofs like this is to translate from set theory concepts to logical concepts, do the essence of the proof in logic, and then translate back to set theory. For example, you can use the following calculation: For every set $A$ and $B$, $$ \begin{align} & B \subseteq A^c \\ \equiv & \;\;\;\;\;\text{"expand definition of $\subseteq$"} \\ & \langle \forall x :: x \in B \;\Rightarrow\; x \in A^c \rangle \\ \equiv & \;\;\;\;\;\text{"expand definition of $\cdot^c$"} \\ & \langle \forall x :: x \in B \;\Rightarrow\; x \not\in A \rangle \\ \equiv & \;\;\;\;\;\text{"logic: expand $\Rightarrow$ -- working towards combining $A$ and $B$ "} \\ & \langle \forall x :: x \not\in B \lor x \not\in A \rangle \\ \equiv & \;\;\;\;\;\text{"logic: rewrite using DeMorgan -- working towards the shape of $A \cap B = \emptyset$"} \\ & \langle \forall x :: x \in A \land x \in B \;\equiv\; \mathrm{false} \rangle \\ \equiv & \;\;\;\;\;\text{"introduce $\cap$ and $\emptyset$ using their definitions"} \\ & \langle \forall x :: x \in A \cap B \;\equiv\; x \in \emptyset \rangle \\ \equiv & \;\;\;\;\;\text{"set extensionality, i.e., definition of $=$ on sets"} \\ & A \cap B = \emptyset \\ \end{align} $$ The key steps are the ones marked 'logic'.

As a nice by-product, we see that we have proved a stronger statement than originally asked: the two expressions $B \subseteq A^c$ and $A \cap B = \emptyset$ are actually equivalent.

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