Boolean Algebra - Prove XYZ + XYZ' + XY'Z + X'YZ = XY + XZ + YZ Trying to prove $((X\land Y\land Z)\lor (X\land Y\land \lnot Z)\lor (X\land \lnot Y\land Z )  \lor (\lnot X\land Y\land Z)) \equiv
 ((X\land Y)\lor (X\land Z)\lor (Y\land Z))$
and I am a bit stuck.
I have the following:
$\equiv X\land Y\land (Z\lor \lnot Z) + Z\land ((X\land \lnot Y)\lor (\lnot X\land Y))\quad$
Distributive   
$\equiv X\land Y\land (1)\lor Z\land((X\land \lnot Y)\lor (\lnot X\land Y))\quad$
Inverse   
$\equiv X\land Y + Z\land(X\land \lnot Y\lor \lnot X\land Y)\quad$    Inverse/Identity    
I'm not sure what needs to be done with the "$Z\land((X\land\lnot Y)\lor (X\land\lnot Y))\quad$" ?
Any help appreciated
 A: Logical terms -- LHS term 1 in this case -- can be used in more than one combination.
LHS terms 1 and 2 combine to yield RHS term 1.
LHS terms 1 and 3 combine to yield RHS term 2.
LHS terms 1 and 4 combine to yield RHS term 3.
A: LHS-
=X'YZ+XY'Z+XYZ'+XYZ
=X'YZ+XY'Z+XY(Z'+Z)
=X'YZ+XY'Z+XY.1
=X'YZ+XY'Z+XY
=X'YZ+X(Y'Z+Y)
=X'YZ+X[(Y'+Y).(Y+Z)]
=X'YZ+X[1.(Y+Z)]
=X'YZ+X(Y+Z)
=X'YZ+XY+XZ
=Y(X'Z+X)+XZ
=Y[(X'+X).(X+Z)]+XZ
=Y[1.(X+Z)]+XZ
=Y(X+Z)+XZ
=XY+YZ+XZ
=RHS.......Hence Proved
A: start with $x + y = xy' + x'y + xy$ 
(proof: $x + y = x1 + 1y = x(y+y') + (x+x')y = xy + xy' + x'y + xy = xy' + x'y + xy$, because $xy + xy = xy$).
So: $xy + xz + yz = xy1 + (x+y)z = xy(z+z') + (x'y + xy' + xy)z = xyz + xyz' + x'yz + xy'z + xyz = xyz + xyz' + x'yz + xy'z.$
A: To add to Kshitiz's answer:

*

*start with: $XYZ+XY'Z+X'YZ+'XYZ$


*apply distributive law: $XY(Z+'Z)+X'YZ+'XYZ$


*apply complement law: $XY1+X'YZ+'XYZ$


*apply identity law: $XY+X'YZ+'XYZ$


*apply distributive law: $X(Y+'YZ)+'XYZ$


*apply absorption law: $X(Y+Z)+'XYZ$


*distribute: $XY+XZ+'XYZ$


*apply distributive law: $XY+Z(X+'XY)$


*apply absorption law: $XY+Z(X+Y)$


*distribute: $XY+ZX+ZY$
https://www.boolean-algebra.com/ is one of the websites which can solve and give steps to solving such problems
