1. Mar 6, 2014

### DopplerFX

I'm trying to simplify 2 boolean expressions. Have I done it correctly and how would I simplify these further if possible. Thank you in advance! :)

Relevant Equations:

(Associativity of +)
A+(B+C) = (A+B)+C​
(Associativity of x)
A*(B*C) = (A*B)*C​
(Commutativity of +)
A+B = B+A​
(Commutativity of x)
A*B = B*A​
(Distributivity of x over +)
A*(B+C) = (A*B)+(A*C)​
(Identity for +)
A+0 = A​
(Identity for x)
A*1 = A​
(Annihilator for x)
A*0 = 0​

(Idempotence of +)
A+A = A​
(Idempotence of x)
A*A = A​
(Absorption 1)
A*(A+B) = A​
(Absorption 2)
A+(A*B) = A​
(Distributivity of + over x)
A+(B*C) = (A+B)*(A+C)​
(Annihilator for +)
A+1 = 1​

This is the first expression:
E = A'B'C'D' + A'B'CD' + A'B'CD + A'BC'D' + A'BC'D + A'BCD' + A'BCD + ABC'D' + ABC'D + ABCD' + ABCD

Here is my working:
E = A'B'C'D' + A'B'CD' + A'B'CD + A'BC'D' + A'BC'D + A'BCD' + A'BCD + ABC'D' + ABC'D + ABCD' + ABCD
E = A'B'C'D' + (D' + D) A'B'C + (D' + D)A'BC' + (D' + D)A'BC + (D' + D)ABC' + (D' + D)ABC
E = A'B'C'D' + (1) A'B'C + (1)A'BC' + (1)A'BC + (1)ABC' + (1)ABC
E = A'B'C'D' + A'B'C + A'BC' + A'BC + ABC' + ABC
E = A'B'C'D' + (B' + B)A'C + (A' + A)BC' + ABC
E = A'B'C'D' + (1)A'C + (1)BC' + ABC
E = A'B'C'D' + A'C + BC' + ABC

This is the second expression:
E = A'B'C'D' + A'B'C'D + A'BC'D' + A'BC'D + A'BCD' + A'BCD

Here is my working:
E = A'B'C'D' + A'B'C'D + A'BC'D' + A'BC'D + A'BCD' + A'BCD
E = B'(A'C'D' + A'C'D) + B(A'C'D' + A'C'D) + A'BCD' + A'BCD
E = (B' + B)(A'C'D' + A'C'D) + A'BCD' + A'BCD
E = (1)(A'C'D' + A'C'D) + A'BCD' + A'BCD
E = (A'C'D' + A'C'D) + A'BCD' + A'BCD
E = A'C'D' + A'C'D + (D' + D)A'BC
E = A'C'D' + A'C'D + (1)A'BC
E = A'C'D' + A'C'D + A'BC
E = (D' + D)A'C' + A'BC
E = (1)A'C' + A'BC
E = A'C' + A'BC
E = A'(C' + BC)

Last edited: Mar 6, 2014
2. Mar 6, 2014

### .Scott

Looks good to me.

3. Mar 6, 2014

### Staff: Mentor

You are not finished yet! This can be further simplified.

The same holds for your earlier working.