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Is this a correct application of the distributive law?

$$\begin{align} (\neg P \wedge \neg Q \wedge R) \vee (\neg P \wedge Q \wedge \neg R) &\equiv \phantom{\wedge}(\neg P \vee \neg P) \wedge (\neg P \vee Q) \wedge (\neg P \vee \neg R) \\ &\phantom{\equiv}\wedge (\neg Q \vee \neg P) \wedge (\neg Q \vee Q) \wedge (\neg Q \vee \neg R) \\ &\phantom{\equiv}\wedge (\phantom{\neg}R \vee \neg P) \wedge (\phantom{\neg}R \vee Q) \wedge (\phantom{\neg}R \vee \neg R) \end{align}$$

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  • $\begingroup$ Hi! Welcome to Math. SE. You should read about truth's table: en.m.wikipedia.org/wiki/Truth_table for verification of your solution. $\endgroup$
    – user798113
    Aug 25, 2020 at 4:03

1 Answer 1

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Yes, you have correctly distributed those two conjunctions.

$${\phantom{\equiv~}(\neg P \wedge \neg Q \wedge R) \vee (\neg P \wedge Q \wedge \neg R)\\ \equiv\\ \phantom{\equiv~}{\phantom{\,\wedge\,}(\neg P \vee \neg P) \wedge (\neg P \vee Q) \wedge (\neg P \vee \neg R) \\\wedge (\neg Q \vee \neg P) \wedge (\neg Q \vee Q) \wedge (\neg Q \vee \neg R) \\\wedge (\phantom{\neg}R \vee \neg P) \wedge (\phantom{\neg}R \vee Q) \wedge (\phantom{\neg}R \vee \neg R)}}$$

However, there is another way to apply distribution: to distribute out the common factor.

$${\phantom{\equiv~}(\neg P \wedge \neg Q \wedge R) \vee (\neg P \wedge Q \wedge \neg R)\\ \equiv\\ \phantom{\equiv~}{\neg P\wedge ((\neg Q\wedge R)\vee(Q\wedge\neg R))}}$$

You can then apply your distribution technique to the right conjunct , and simplify the expression in a few more steps:

$${\phantom{\equiv~}\neg P\wedge ((\neg Q\wedge R)\vee(Q\wedge\neg R))\\\equiv\\ \phantom{\equiv~}{\neg P\wedge ((\neg Q\vee Q)\wedge(\neg Q\vee\neg R)\wedge (R\vee Q)\wedge(R\vee \neg R))}\\\equiv\\~~~\ddots}$$

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