# Simplify boolean expression X'YZ + XY'Z + XYZ'

i have this expression to put in the XOR forms.

X'YZ + XY'Z + XYZ'


The steps i did already are these ones:

Z(X'Y + XY') + XYZ'
Z(X^Y) + XYZ'


But if i put the same expression on WolframAlpha it says that the final solution should be:

XY ^ XZ ^ YZ ^ XYZ

i know that AB' + A'B = A^B, but in this case i dont know how X^Y can be correlated to XY and obtain the full solution.

Can someone help me?

Thank you

## 1 Answer

\begin{eqnarray*} XY \hat{ } XZ \hat{ } YZ \hat{ } XYZ & = & (XY (XZ)^{'}+ (XY)^{'}XZ) \hat{ }(YZ (XYZ)^{'}+ (YZ)^{'}XYZ) \\ & = & (XY (X^{'}+Z^{'})+ (X^{'}+Y^{'})XZ) \hat{ } (YZ (X^{'}+Y^{'}+Z^{'})+ (Y^{'}+Z^{'})XYZ) \\ & = & (XY Z^{'}+ Y^{'}XZ)\hat{ }(YZ X^{'}) \\ & = & (XY Z^{'}+ Y^{'}XZ)^{'}(X^{'} YZ )+(XY Z^{'}+ Y^{'}XZ)(YZ X^{'})^{'} \\ & = & ((XY Z^{'})^{'} (Y^{'}XZ)^{'})(X^{'} YZ )+(XY Z^{'}+ Y^{'}XZ)(Y^{'}+Z^{'} +X) \\ & = & (X^{'}+Y^{'}+ Z) (X^{'}+Y+Z^{'})(X^{'} YZ )+XY Z^{'}+ Y^{'}XZ \\ & = & X^{'} YZ +XY Z^{'}+ Y^{'}XZ \\ \end{eqnarray*}

• Really thanks, but how can you write X'YZ as (X′+Y′+Z)(X′+Y+Z′)(X′YZ)?? – MatPag Mar 20 '17 at 15:43
• Expand out the nine terms, $5$ will be killed by $b b^{'}=0$, leaving the 4 terms $X^{'}X^{'}X^{'}YZ+X^{'}YX^{'}YZ+ZX^{'}X^{'}YZ+ZYX^{'}YZ$, these are all equal to $X^{'}YZ$, then use $a+a=a$. This was a sizeable leap. ... Problems like this are probably most easily verified by using a truth table ... Boolean algebra is tough (espacially if you are required to state every axiom that is used !) – Donald Splutterwit Mar 20 '17 at 20:17