- #1
DopplerFX
- 2
- 0
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 +)
(Idempotence of +)
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)
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)
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