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Use the laws of sets to prove:

a. $ C- [A^{c} ∪ (A ∪ B)^{c}]^{c} = C ∩ A^{c} $

I am struggling regarding how to prove this.

De Morgan's Law: $ C- [A^{c} ∪ (A^{c} ∩ B^{c})]^{c} $

Distributive Law: $ C- [(A^{c} ∪ A^{c}) ∩ (A^{c} ∪ B^{c})]^{c} $

Idempotent Law: $ (A^{c} ∪ A^{c}) = A^{c} $

$ C- [(A^{c}) ∩ (A^{c} ∪ B^{c})]^{c} $

Are my workings so far correct? I would appreciate any assistance in how to proceed in proving this.

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  • $\begingroup$ It would become $C- [A^{c}]^{c} \; = \; C- A \; = \; C ∩ A^{c}$ $\endgroup$ – L KM Feb 18 '18 at 12:29
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You did well so far. As L KM wrote in the comment, you can use absorption law $X\cap (X\cup Y)=X$ and choose $X=A^C$ and $Y=B^C$. Then, you are almost done.

Remark: You can do is even shorter. Use the other absorption law $X\cup (X\cap Y)=X$ with $X=A^C$ and $Y=B^C$ right after your first step.

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You need to use the definition of set minus:

$$A - B =_{df} A \cap B^C$$

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