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$A ∩ B = A ∩ (B' ∩ A)'$ if $A,B ⊆ U$ (universal set)

so to prove it I need to show that both sides are a subset of each other. I did the right side in the following steps:

  1. De Morrigan: $A ∩ (B' ∩ A)' = A ∩ (B'' ∪ A') $

  2. Universal set laws: $A ∩ (B'' ∪ A') = A ∩ (B ∪ A')$

I could have used distribution law but based on how they are not allowing us to use them so they can make each homework harder I went with:

  1. Intersection def: $∀x∈ A ∩ (B ∪ A'), x∈A \,\text{&}\, x∈(B ∪ A')$

  2. $... x∈A \,\text{&}\, x∈B \,\text{or}\, x∈A'$

  3. Universal set laws: $x∈A\,\text{ &}\, x∈B \,\text{or}\, x∉A$

  4. if $x∉A$ it a contradiction with $x∈A$ which means that $x∈B$ is the correct one.

  5. thusly we arrive at $x∈A$ & $x∈B \implies x∈A ∩ B \implies A ∩ (B' ∩ A)' ⊆ A ∩ B$.

now for the other side, I tried everything, and the best thing I came out with is:

$A ∩ B \implies A ∩ B ∪ ∅ \implies$

$A ∩ B ∪ ( A ∩ A') \implies A ∩ [B ∪ ( A ∩ A')] \implies$

$A ∩ [(B ∪ A) ∩ ( B ∪ A')] \implies A ∩ [(B ∪ A) ∩ ( B ∪ A')]'' \implies$

$[(B ∪ A)' ∪ ( B ∪ A')']' \implies A ∩ [(B' ∩ A') ∪ ( B' ∩ A)]'$ and from here I spiraled in many ways etc.

SIDENOTE: thank you for your advice regarding deleting the original post and giving more context, it was my first time using the site and I posted it out of desperation at like 5 pm my time.

Asaf Karagila
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Abdallh Bder
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1 Answers1

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Assume $x\in A\cap B \implies x\in A $ and $x\in B$. You've already shown that $A\cap (B'\cup A)'=A\cap(B\cup A'). $

Therefore, since $x\in B$, we know that $x\in B\cup A'$. We already know $x\in A$ and therefore, $x\in A \cap (B\cup A')=A\cap (B'\cup A)'$ so $A \cap B\subseteq A\cap (B'\cup A)'$.

A P
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