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I am going through the accepted proof listed in a previous thread and got stuck on a step where you use the distributivity property to go from:

(¬P∧R)∨(¬P∧¬Q)∨(R∧Q)

to:

(¬P∨¬P∨R)∧(¬P∨¬Q∨R)∧(¬P∨¬P∨Q)∧(¬P∨¬Q∨Q)∧(R∨¬P∨R)∧(R∨¬Q∨R)∧(R‌​∨¬P∨Q)∧(R∨¬Q∨Q)

Unfortunately, I was not able to understand how this was done.

Could someone show me the steps of how to do this?

tree_traversal
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1 Answers1

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We just use the distributive law:

$$a\lor(b \land c)\equiv(a \lor b)\land(a \lor c)$$

Distributivity is analogous to distributivity of multiplication over addition: $$a \times (b + c) = a\times b + a\times c$$

For eg. in your case, we have

$$=(a \land b) \lor (c \land d) \lor (e \land f)$$ $$\equiv (a + b) \times (c + d) \times (e + f)$$ (consider symbolic change for understanding)

$$\equiv (a \times c \times e) + (a \times c \times f) + (a \times d \times e)+ (a \times d \times f) + (b \times c \times e) + (b \times c \times f)+ (b \times d \times e)+ (b \times d \times f)$$

which is exactly what is written as $$(\neg P \lor \neg P \lor R) \land (\neg P \lor \neg Q \lor R) \land (\neg P \lor \neg P \lor Q) \land (\neg P \lor \neg Q \lor Q) \land (R \lor \neg P \lor R) \land (R \lor \neg Q \lor R) \land (R \lor \neg P \lor Q) \land (R \lor \neg Q \lor Q)$$

ab123
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