- #1

s3a

- 818

- 8

The problem I need help with is the following.:

## Homework Statement

"Simplify to obtain minimum SOP.

F(A, B, C, D) = A’B’CD’+AC’D’+ABC’+AB’C+AB’C+BC’D"

The problem stated above has two provided solutions, the "main" one and the "alternate" one.

I'm confused with both of them.

__Here they are (within this post directly and indirectly within a PDF, since some people prefer it directly here, but I think it looks better in the pdf).:__

"Main" solution:

"Main" solution:

("*"s are used for preserving formatting)

A’B’CD’ + AC’D’ + ABC’ + AB’C + AB’C + BC’D

*2*********8,12****12,13**10,11**10,11***5,13

...*******************ABC'(D+D')*************ABC' is deleted (There is an arrow going from AC'D' on the line above to the left whitespace of ABC'(D+D'), on this line. There is also another arrow going from BC'D to ABC'(D+D') (from the left - not that I think that the direction from which it's coming matters).)

2 & 10,11**A'B'CD' + AB'C = B'C(A+A'D)

Final result: B'CD' + AC'D' + AB'C + BC'D

__"Alternate" solution:__

= A’B’CD’ + AC’D’ + ABC’ +AB’C + AB’C + BC’D

= A’B’CD’ + (C’D’ + BC’ + B’C)A + BC’D

= A(CC’+B’C) + BC’D + A’B’CD’

= AC’D’ +A’B’CD’ +AB’C +BC’D

= AC’D’ + B’CD’+AB’C + BC’D

__PDF version of "main" and "alternate" solutions (The problem is problem 1a.):__

https://www.docdroid.net/PacYXo3/1a-main-solution-and-alternate-solution.pdf

## Homework Equations

A + 1 = 1 Annulment

A + 0 = A Identity

A + 0 = A Identity

A ⋅ 1 = A Identity

A ⋅ 0 = 0 Annulment

A + A = A Idempotent

A ⋅ A = A Idempotent

(A')' Double Negation

A + A' = 1 Complement

A ⋅ A' = 0 Complement

A+B = B+A Commutative

A⋅B = B⋅A Commutative

(A+B)' = A' ⋅ B' de Morgan’s Theorem

(A⋅B)' = A' + B' de Morgan’s Theorem

Source:

https://www.electronics-tutorials.ws/boolean/bool_6.html

## The Attempt at a Solution

When I tried to follow the logic of the solutions, I encountered some problems.

__Here they are.:__

For the main solution of Q1 a):

For the main solution of Q1 a):

1)

What are the numbers 2, 8, 12, 12, 13, 10,11, 10,11, 5, 13?

2)

What is meant by ABC’ is deleted? Is what is meant (D + D’) is “deleted”?

3)

What is done on the before-last linute to get to the last line / line with text “Final Result:”?

__For the alternate solution of Q1 a):__

4)

How does one go from (C’D’ + BC’ + B’C)A to A(CC’+B’C)?

5)

How does one go from A(CC’+B’C) to AC’D’ +AB’C?

6)

How does the A’B’CD’ become B’CD’ from the before-last line to the last line? Any input would be greatly appreciated!