Boolean Equation Simplification

[Berbanog]

Homework Statement

Q1: Show that
a) ~(~(A+B)~C.A).~(~(B.C)+A.B) = ~(~(C.~A)+~B.C)~(A.~B.~(C+~B))
a-2) *same as above* ~(~(A+B).C'.A).~(~(B.C)+A.B) = ~(~(C.A')+B'.C).~(A.B'.~(C+B'))
b) ~(A XOR B) = (~A) XOR B
c) A + A'B = A + B

Q2: Represent the following using only NAND gates, and only NOR gates
a) A.B + ~(A.C).~(B+C)
b) (A XOR B) + ~(A XOR B).(B + C)
c) (A+B).(~(A+C)+~(B.C))

Homework Equations

The Attempt at a Solution

a)
LHS = A + (~A)BC = A + BC
RHS = ~A + BC
*where did i do wrong?
b) Tried Truth Table
A B || ~(A XOR B)
---------------------
0 0 || 1
0 1 || 0
1 0 || 0
1 1 || 1

A B ~A|| (~A) XOR B
---------------------
0 0 1 || 0
0 1 1 || 1
1 0 0 || 1
1 1 0 || 0

Is this suppose to be correct? i find it confusing...to me
(~A) XOR B = A XOR B...but ofcourse the question is asking to prove
~(A XOR B) = (~A) XOR B!

c) I'm quite sure this was distributive law...but how do i prove this WITHOUT using Karnaugh map?

Q2

I : Trial and error solved the question backward!
II: I first simplied the question using theorems & K-map, then with the result figured out the NOR/NAND equation.
a) i got 2 solutions on this one..(?!)
I : ~(~A+~B~(~A+~C)+~(~B.~C)) or in different form...~(A'+B'~(A'+C')+~(B'C'))
II: ~(A~BC) or in different format...~(AB'C)

Method used
b) don't know where i have to begin this since i ain't that great with XOR complements...If someone could give me any tips i'd appreciate it!
c) ~(~A+B(~B)+C) or in different form...~(A'+BB'+C)

 

[KataKoniK]

Thank you for your post. I would like to provide some guidance and suggestions for solving these problems:

Q1:
a) For the first problem, it seems that your attempt is correct. However, it may be helpful to show your steps in simplifying the expressions. For example, you could start by using De Morgan's laws to rewrite the expressions in terms of AND, OR, and NOT operations. Then, you could use distribution and simplification rules to simplify the expressions further. Finally, you could show that the two sides are equivalent by using substitution and logical equivalences.

b) Your approach using a truth table is a good start. However, it may be helpful to also use logical equivalences to show that ~(A XOR B) is equivalent to (~A) XOR B. For example, you could use the following steps:
~(A XOR B)
= ~(A'B + AB')
= (~A + B)(A + ~B)
= (~A + B) XOR (A + ~B)
= (~A XOR A) OR (~A XOR ~B) OR (B XOR A) OR (B XOR ~B)
= 1 OR (~A XOR ~B) OR (B XOR A) OR 0
= (~A XOR ~B) OR (B XOR A)
= (~A XOR B) OR (A XOR ~B)
= (~A XOR B) XOR (A XOR ~B)
= (~A) XOR B

c) For this problem, you could start by using distribution and simplification rules to rewrite the expressions in terms of AND, OR, and NOT operations. Then, you could use logical equivalences to show that the two sides are equivalent.

Q2:
a) For this problem, you could start by simplifying the expression using distribution and simplification rules. Then, you could use logical equivalences to show that the two sides are equivalent. As for your two solutions, both seem to be correct. They are just different ways of representing the same expression.

b) For this problem, you could start by using logical equivalences to rewrite the expression in terms of AND, OR, and NOT operations. Then, you could use De Morgan's laws to rewrite the expression in terms of NAND and NOR operations. Finally, you could use logical equivalences to show that the two sides are equivalent.

c) For this problem, you

 

[Mene]

OR ~(A~B+C) or in different form...~(AB'C)

Hello! I am a scientist and I would like to provide a response to your content on Boolean equation simplification.

Firstly, for Q1a, you have correctly simplified the left-hand side (LHS) to A + BC. However, for the right-hand side (RHS), it should be ~A + BC. This is because ~(C.~A) = A + C, and ~(C+~B) = ~B + C. By substituting these in, we get ~(A + ~B + C) which simplifies to ~A + BC. Therefore, the LHS and RHS are equivalent.

For Q1b, you are correct in using a truth table to prove that ~(A XOR B) is equivalent to (~A) XOR B. Alternatively, you can also use De Morgan's law to simplify ~(A XOR B) to (~A) AND (~B), which is equivalent to (~A) XOR B.

For Q1c, you can use the distributive law to prove that A + A'B is equivalent to A + B. This can be done by expanding A'B to (A+B)(A+C) and simplifying it to B. Therefore, A + A'B is equivalent to A + B.

Moving on to Q2, for part a, you have correctly simplified the equation using De Morgan's law. For part b, you can use De Morgan's law again to simplify ~(A XOR B) to (~A) AND (~B), and then use the distributive law to distribute the AND operator over the OR operator. This will result in (~A) OR (~B) OR (~A)(~C) OR (~B)(~C). You can then use De Morgan's law again to simplify this to ~(A AND B) OR (~(A OR B) AND ~C). This is equivalent to the given equation, (A XOR B) + ~(A XOR B)(B + C).

For part c, you can use De Morgan's law to simplify ~(A+B) to (~A) AND (~B). Then, you can use the distributive law to distribute the AND operator over the OR operator, resulting in (~A) OR (~B) OR (~A)(~C) OR (~B)(~C). This is equivalent to the given equation, (A