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I am in the process of converting a boolean formula into disjunctive normal form. I have reached a point where the formula looks like this:

(X∨Y) ∧ (X∨¬Y)

Does the distributive law of ∧ over ∨ allow the formula to be this in the next step of the conversion?:

(X∧X) ∨ (X∧¬Y) ∨ (Y∧X) ∨ (Y∧¬Y)

i.e. have I used the distributive law correctly here?

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    Why not apply [Distributivity](https://en.wikipedia.org/wiki/Distributive_property#Rule_of_replacement) to get : $X \lor (Y \land \lnot Y) \equiv X$ ? – 2017-01-16
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    Your rewriting is _valid_ and would be the way to proceed towards a DNF in the mindless follow-the-recipe fashion -- but using distributivity in the direction @Mauro suggests will be more _useful_ for most purposes when doing things by hand. – 2017-01-16
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    Okay thank you for your help, I will remember to use this – 2017-01-16

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When doing boolean algebra it is common to encounter a formula like yours, and because it simplifies to something much smaller, it has been given a name:

Adjacency

$(X \lor Y) \land (X \lor \neg Y) \Leftrightarrow X$

$(X \land Y) \lor (X \land \neg Y) \Leftrightarrow X$

Remember this equivalence so you can spot it when it occurs and simplify it in one immediate step, rather than making things more complicated, as you were originally doing.

By the way, the name 'adjacency' comes from the fact that if you were to do a Karnaugh Map (K-Map) for the left hand side, you would find that the two terms are adjacent in the map, and can be captured by the more simpler term on the right hand side.

Oh, and here are two more pairs of equivalences that are really useful, but that people tend to forget or overlook:

Absorption

$P \land (P \lor Q) \Leftrightarrow P$

$P \lor (P \land Q) \Leftrightarrow P$

Reduction

$P \land (\neg P \lor Q) \Leftrightarrow P \land Q$

$P \lor (\neg P \land Q) \Leftrightarrow P \lor Q$

Add these three pairs of equivalences to your 'boolean algebra toolbox'!