Boolean Algebra - Prove XYZ + XYZ' + XY'Z + X'YZ = XY + XZ + YZ

Question

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

Answer 1

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: $xy + xz + yz = xy1 + (x+y)z = xy(z+z') + (x'y + xy' + xy)z = xyz + xyz' + x'yz + xy'z + xyz = xyz + xyz' + x'yz + xy'z.$

Answer 2

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.

Answer 3

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

Answer 4

To add to Kshitiz's answer:

1. start with: $XYZ+XY'Z+X'YZ+'XYZ$

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

3. apply complement law: $XY1+X'YZ+'XYZ$

4. apply identity law: $XY+X'YZ+'XYZ$

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

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

7. distribute: $XY+XZ+'XYZ$

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

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

10. distribute: $XY+ZX+ZY$

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