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.

I did not downvote your problem, but it is not very understandable as written. You should say up front that $A$, $B$, and so on are Boolean variables. — kimchi lover, Sep 30 '19 at 00:36
@IntegrateThis Yes i am looking forward to answer thank you for edits and incoming answer i appreciate the help — Zs11601, Sep 30 '19 at 00:49

score 1 · Answer 1 · answered Sep 30 '19 at 00:58

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*}

answered Sep 30 '19 at 00:58

Adriano

41,576

Sir i dont understand domination law – Zs11601 Sep 30 '19 at 01:03
1

Is it just that 1+anything is 1? – Zs11601 Sep 30 '19 at 01:04
Thank you for solution! – Zs11601 Sep 30 '19 at 01:30

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

1 Answers1