How to reduce A'B'+AB'+BC'D' to simplest form?

Question

I wish to reduce the expression $A'B'+AB'+BC'D'$ to its simplest form.

The notation for boolean variables $A,B,C,D$, for those who are not familiar, is :

$+$ for logical OR operation

$'$ for the logical NOT operation, and

$AB$ means $A$ logical AND $B$

The answer is B’ + C’D’, and my attempt at the solution is:

$A'B'(B+B') +AB'+BC'D'$

$\equiv AB'B+A'B'B'+AB'+BC'D'$

$\equiv A'+AB'+A'B'+BC'D'$

$\equiv A'+B'+A'B'+BC'D'$

Here I am stuck. Any help much appreciated.

Answer 1

Observe that: \begin{align*} A'B' + AB' + BC'D' &= (A' + A)B' + BC'D' & \text{Distributive Law} \\ &= (1)B' + BC'D' & \text{Inverse Law} \\ &= (1 + C'D')B' + BC'D' & \text{Domination Law} \\ &= B' + B'C'D' + BC'D' & \text{Distributive Law} \\ &= B' + (B' + B)C'D' & \text{Distributive Law} \\ &= B' + (1)C'D' & \text{Inverse Law} \\ &= B' + C'D' & \text{Identity Law} \\ \end{align*}