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

\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*}
