Proving $(Y\cap Z)\cup (X \cap Z ) \cup (X \cap Y )= ((Y \cup Z) − (\bar{X} \cap \bar{Z})) \cap (X \cup Y)$

Question

I'm trying to prove that $$(Y\cap Z)\cup (X \cap Z ) \cup ({X} \cap {Y} )= ((Y \cup Z) − (\bar{X} \cap \bar{Z})) \cap (X \cup Y)$$ using set identities. I have been using mainly de morgans law and gotten the left side to become $$\overline{((\bar{Y}\cup \bar{Z}) - (X \cap Z )) \cap (\bar{X} \cup \bar{Y})}$$ but I can't quite get to to become the right side after a lot of efford.

Answer 1

In general I find it easier to start on the more complicated side, which in this case is the righthand side. I’ll let $U$ be a universal set for the problem. First,

$$\begin{align*} (Y\cup Z)\setminus\big((U\setminus X)\cap(U\setminus Z)\big)&=(Y\cup Z)\setminus\big(U\setminus(X\cup Z)\big)\\ &=(Y\cup Z)\cap(X\cup Z)\;, \end{align*}$$

so the righthand side reduces to

$$(Y\cup Z)\cap(X\cup Z)\cap(X\cup Y)\;.$$

Apply a distributive law to pull $Z$ out of the first two terms to get

$$\big((Y\cap X)\cup Z\big)\cap(X\cup Y)\;,$$

and then distribute the final $X\cup Y$ through the first term:

$$\big((Y\cap X)\cap(X\cup Y)\big)\cup\big(Z\cap(X\cup Y)\big)\;.$$

Now use the fact that $Y\cap X\subseteq X\cup Y$, and hence $(Y\cap X)\cap(X\cup Y)=Y\cap X$, to get

$$(Y\cap X)\cup\big(Z\cap(X\cup Y)\big)\;.$$

(Depending on just what identities you have available, this may require more than a single step.) One last application of a distributive law essentially finishes it off:

$$(Y\cap X)\cup\big(Z\cap(X\cup Y)\big)=(Y\cap X)\cup(Z\cap X)\cup(Z\cap Y)\;.$$