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I have the following logical expression to solve/Simplify. Anyone able to help please??

$(P\land Q\land R)\lor (P\land\lnot Q\land R)\lor (P\land \lnot Q\land \lnot R)$

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thanks @Asaf -- – user10695 May 9 '11 at 19:24
up vote 5 down vote accepted

We proceed to employ the distributive laws of $\land$ and $\lor$ over one another, DeMorgan's law about negation, and the associativity of these operations.

Here it gets a bit technical:

$$\begin{align} &(P\land Q\land R)\lor (P\land\lnot Q\land R)\lor (P\land \lnot Q\land \lnot R) &\iff\\ &(P\land Q\land R)\lor (P\land \lnot Q\land R)\lor (P\land \lnot(Q\lor R))&\iff\\ &(P\land Q\land R)\lor (P\land \lnot(Q\lor\lnot R))\lor (P\land \lnot(Q\lor R))&\iff\\ &(P\land (Q\land R))\lor (P\land (\lnot(Q\lor\lnot R)\lor \lnot(Q\lor R)))&\iff\\ &P\land ((Q\land R)\lor (\lnot(Q\lor\lnot R)\lor \lnot(Q\lor R)))&\iff\\ &P\land ((Q\land R)\lor\lnot((Q\lor\lnot R)\land (Q\lor R)))&\iff\\ &P\land ((Q\land R)\lor\lnot(Q\lor (\lnot R\land R)))&\iff\\ &P\land ((Q\land R)\lor\lnot Q)&\iff\\ &P\land ((\lnot Q\lor Q)\land (\lnot Q\lor R))&\iff\\ &P\land (\lnot Q\lor R) \end{align}$$

This seems simplified enough.

Addendum: Another way to solve such problem (especially when number of variables is very low) is to write a very big truth table, in this case, 8 lines long.

Then you can look closely and see when the sentence is true, it will happen to be true when $P$ is true, and either $Q$ is false or $R$ is true, this means that the original sentence is equivalent to this very short one.

This method though is very hard to do when there are more than three variables, or when the original sentence is very very complicated.

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