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I am told to use the laws of set algebra to simplify $(A \cap B^c) \cup (A^c \cap B^c)^c$.

And I need to show the reason for each step.

$(A \cap B^c) \cup (A^c \cap B^c)^c$

$= (A \cap B^c) \cup (A \cup B)$ By De Morgan's law.

$= (B^c \cap A) \cup (A \cup B)$ By commutativity.

$= B^c \cap (A \cup A) \cup B$ By associativity.

$= B^c \cap A \cup B$ By idempotence.

$= B^c \cap B \cup A$ by commutativity

$= \emptyset \cup A$ Since we have intersection with complement.

$= A$ By identity laws.

Does this all look correct? If not then what is the correct way? Thanks for any help.

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  • $\begingroup$ What happens if A is false and B is true? I would re-examine the last four lines. $\endgroup$ – John Douma Aug 26 '18 at 19:23
  • $\begingroup$ @JohnDouma What is wrong with it? I do not see anything? $\endgroup$ – Wyuw Aug 26 '18 at 19:30
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    $\begingroup$ The third line with the = sign is incorrect. You cannot use associativity there. $\endgroup$ – John Douma Aug 26 '18 at 19:33
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    $\begingroup$ $(B^\complement\cap A)\cup A\cup B=((B^\complement \cap A)\cup A)\cup B$ by associativity. Next use absorption... $\endgroup$ – Graham Kemp Aug 26 '18 at 21:41
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    $\begingroup$ Yes. You cannot make it any simpler than that, @Wyuw. $\endgroup$ – Graham Kemp Aug 26 '18 at 23:38
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Note that at $$(B^c \cap A) \cup (A \cup B)$$ we have $$(B^c \cap A) \subseteq (A \cup B) $$

Therefore the answer should be $$(B^c \cap A) \cup (A \cup B)= (A \cup B)$$

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Line 3 to 4 is the issue, ie $(B^c \cap A) \cup (A \cup B) \neq B^c \cap (A \cup A) \cup B$.

For any four sets $A, B, C$ and $D$ associativity implies that $(A \cap B) \cup (C \cup D) = ((A \cap B) \cup C) \cup ((A \cap B) \cup D)$ so in this case we have that $$ (B^c \cap A) \cup (A \cup B) = ((B^c \cap A) \cup A) \cup ((B^c \cap A) \cup B)$$

But this step is not necessary since as stated by @Mohammad Riazi-Kermani, you just need to notice that $B^c \cap A \subseteq A$ and therefore $(B^c \cap A) \cup (A \cup B) = A \cup B$

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