Simplify the Boolean expression $X Y Z+X Y' Z'+X' Y' Z+X' Y Z'$?

Question

$$X Y Z+X Y' Z'+X' Y' Z+X' Y Z'$$

I know it simplify to (X XOR Y XOR Z),BUT I want to simplified using only AND, OR, and NOT Gates?

Please help I spent three hours but I don't get the same truth table.

Answer 1

I suppose the prime means negation and then $X'=1+X$ etc.

$XYZ+XY'Z'+X'Y'Z+X'YZ'=$ $XYZ+X(1+Y)(1+Z)+(1+X)(1+Y)Z+(1+X)Y(1+Z)=$ $XYZ+X(1+Y+Z+YZ)+(1+X+Y+XY)Z+(Y+XY)(1+Z)=$ $XYZ+X+XY+XZ+XYZ+Z+XZ+YZ+XYZ+Y+YZ+XY+XYZ=$ $X+Y+Z$

Since $X+Y=(X\vee Y)\wedge \neg(X\wedge Y)$ just use AND, OR and NOT to construct $X+Y+Z$, that is $X$ XOR ($Y$ XOR $Z$)

Answer 2

Let T : C3 → C3 be the linear operator defined by T(z1, z2, z3) = (z1 + iz2, iz2 - z3, z1 + z3). (1.1) Find a formula for T ∗(z1, z2, z3) expressed as a single vector in terms of z1, z2, z3. (1.2) Find a basis for N (T) . (1.3) Find a basis for R (T ∗) . (1.4) Show that R (T ∗)⊥ = N (T) .