Simplify using the set rules of inference

**aaronrj** · February 22nd 2010, 06:26 PM

Simplify the following: A, B, and C are subsets of the Universe.

$\overline{[((A-B)\cup(A \cap C)) \cap\overline{(A \cap B)}}$

**Soroban** · February 22nd 2010, 10:28 PM

Hello, aaronrj!

I think I've got it . . .

$A, B,C$ are subsets of the Universe.

Simplify: . $\overline{\left[\,(A-B)\cup(A \cap C)\,\right] \:\cap \:\overline{(A \cap B)}}$

Under the long negation bar, we have:

$\left[\,(A - B) \cup (A \cap C)\,\right] \:\cap \:\overline{(A \cap B)}$

. . $=\; \left[\,(A \cap \overline{B}) \cup (A \cap C)\,\right] \:\cap\:(\overline{A} \cup \overline{B})$ . . Def. of set subtraction, DeMorgan's Law

. . $=\;\left[\,A \cap (\overline{B} \cup C)\,\right] \cap (\overline{A} \cup \overline{B})$ . . . . . . Distributive Property

. . $=\;A \cap \left[\,(\overline{B} \cup C) \cap (\overline{A} \cup \overline{B})\,\right]$ . . . . . . Associative Property

. . $=\; A \cap \left[\,(\overline{B} \cup C) \cap (\overline{B} \cup \overline{A})\,\right]$ . . . . . . Commutative Property

. . $=\;A \cap \left[\,\overline{B} \cup (C \cap \overline{A})\,\right]$ . . . . . . . . . .Distributive Property

. . $=\; (A \cap \overline{B}) \,\cup \, \left[\,A \cap (C \cap \overline{A})\,\right]$ . . . . . .Distributive Property

. . $=\;(A \cap \overline{B}) \,\cup\,\underbrace{\left[\,A \cap \overline{A} \cap C\,\right]}_{\text{This is false}}$ . . . . . . .Comm., Assoc. Properties

. . $=\qquad A \cap \overline{B}$ . . . . . . . . . . . . . . .Don't know the name of this property.

Replacing the long negation bar, we have:

. . . . $\overline{(A \cap \overline{B})} \;\;=\;\;\overline{A} \cup B$ . . . . . . . . . DeMorgan's Law

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