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Can I simplify

$(F \land \neg M ) \lor (F \land A) \lor (F \land M \land G)$

to

$F \land (\neg M \lor A \lor (M \land G) )$

? And if so, what law does this illustrate?

Baodad
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2 Answers2

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The distributive law is the factoring law. You have factorised the expression by taking out the common factor of F.

So it is the distributive law.

Dan
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    Over at Wikipedia, I'm reading about two kinds of distributivity: Distributivity of $\land$ over $\lor$ (similar to ordinary algebra), and distributivity of $\lor$ over $\land$ (not like ordinary algebra). Which one of these (or both) are in use in this OP? – Baodad Nov 29 '16 at 17:34
  • Both are in use. – Dan Nov 29 '16 at 17:37
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$(\color{blue}{F} \land \neg M ) \lor (\color{blue}{F} \land A) \lor (\color{blue}{F} \land M \land G)$

We can use the distributive law to get:

$\color{blue}{F} \land (\neg M \lor A \lor (M \land G) )$ You can simplify by first using commutativity of disjunction to obtain $$F \land( A \lor\lnot M \lor(M\land G))$$ and then $$F\land (A \lor((\lnot M\lor M) \land (\lnot M \lor G))$$

Which can be simplified to

$$F\land (A \lor \lnot M \lor G))$$

amWhy
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  • The transition from top expression, using distributivity of And over Or. Then after commuting I distribute Or over And – amWhy Nov 29 '16 at 17:47