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From my textbook, by using distributive law, its able to simplify:

$[(p \land \lnot q) \lor (p \land q)] \land q$

To:

$[p \land (\lnot q \lor q)] \land q $

I don't know how to get to this step, and here is how I've tried by distributing the first expression :

$[(p \lor p) \land (p \lor q) \land (\lnot q \lor p) \land (\lnot q \lor q)] \land q$

If I try continue to expand this, it will become very complex. I think I'm doing this in the wrong way and I will appreciate it very much if anyone can explain how the textbook got the simplified answer.

Thanks

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2 Answers 2

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$$[(\color{blue}{p} \land \lnot q) \lor (\color{blue}{p} \land q)] \land q \equiv [\color{blue}{p} \land (\color{red}{{\lnot q \lor q}})] \land q\tag{distributivity}$$

$$ \equiv (\color{blue}{p} \land \color{red} \top) \land q$$

$$\equiv \color{blue}{p} \land q$$

That is, the distributive law declares that $$[p\land (\lnot q \lor q)] \equiv [(p \land \lnot q) \lor (p \land q)]$$ That means that it is the same law when we write it: $$[(p \land \lnot q) \lor (p \land q)] \equiv [p\land (\lnot q \lor q)]$$


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The Distributive law goes both ways, because it is an equivalence.

That is, using the Distributive Law you can go from $p \land (\neg q \lor q)$ to $(p \land \neg q) \lor (p \land q)$, but you can use that same Distributive Law to go from the latter to the former, and that is what they did here.

I believe the name is somewhat to blame for the fact that so many students, like you, only think of the Distributive Law going one way, because going from $(p \land \neg q) \lor (p \land q)$ to $p \land (\neg q \lor q)$ does not feel like Distribution, but rather as a kind of 'Un-Distribution' or 'Reverse Distribution', or maybe even 'Taking out a common Factor' ... but it is still called Distribution in logic.

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  • $\begingroup$ @Bram28 ahh I see, thank you. I've tried to test the result on truth table, and the two expression are indeed equivalent, however, is there a way to prove their equivalency without using truth table? Is it possible to use the distributive law on [(p∧¬q)∨(p∧q)] and arrive at [p∧(¬q∨q)] ? It seems it just goes on forever with the expansion. $\endgroup$ Sep 9, 2018 at 0:07
  • $\begingroup$ @IhavelowIQ Yes. This is what I tried to say in my Answer: going from $[(p \land \neg q) \lor (p \land q)]$ to $[p \land (\neg q \lor q)]$ is using the Distributive Law. $\endgroup$
    – Bram28
    Sep 9, 2018 at 0:11
  • $\begingroup$ @IhavelowIQ All these laws in Boolean algebra go both ways. For example, using DeMorgan you can go from $\neg (p \lor q)$ to $\neg p \land \neg q$, but going from $\neg p \land \neg q$ to $\neg (p \lor q)$ is also using the DeMorgan's law, since the DeMorgan's law is an equivalence. So you can go back and forth using the same law. Same for Distribution. $\endgroup$
    – Bram28
    Sep 9, 2018 at 0:14
  • $\begingroup$ Oh sorry, I missed your edited message. You are right, I was only thinking of going one way (taught in the book), and [(p∧¬q)∨(p∧q)] to [p∧(¬q∨q)] was definitely not apparent for me. But thanks for clarifying this. $\endgroup$ Sep 9, 2018 at 0:15
  • $\begingroup$ @IhavelowIQ ok, glad I could help! :) $\endgroup$
    – Bram28
    Sep 9, 2018 at 0:16

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