Looking for steps/theorems for this Boolean algebra conversion

Question

I'm usually pretty good at Boolean algebra, but I can't seem to figure out the theorems/steps involved in the following equation conversion:

C = AB + AD + BD

to

C = AB + D(A xor B)

Any help is appreciated. Thanks

Answer 1

$$AD + BD = D(A+B)\qquad\text{distribution}$$ $$D(A+B) = D(AB + \overline{A}B + A\overline{B})\qquad\text{showing all cases for OR}$$

$$AB + D(AB + \overline{A}B + A\overline{B}) = AB + DAB + D(\overline{A}B+A\overline{B})\qquad\text{distribution}$$

$$AB + DAB + D(\overline{A}B+A\overline{B}) = AB + D(\overline{A}B+A\overline{B})\qquad\text{absorption: }X+XY = X$$

$$AB + D(\overline{A}B+A\overline{B}) = AB + D(A\oplus B)\qquad\text{definition of xor}$$

Answer 2

Here is a useful equivalence:

Reduction

$$P + P'Q = P + Q$$

which can be generalized to:

Generalized Reduction

$$PR + P'QR = PR + QR$$

Applied to your statement:

$$AB + D(A \oplus B) \overset{Def. \ \oplus}{=}$$

$$AB + D(AB' + A'B) \overset{Distribution}{=}$$

$$AB + DAB' + DA'B \overset{Gen'd Reduction \times 2}{=}$$

$$AB + DA + DB$$