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Questions tagged [boolean-algebra]

Boolean algebras are structures which behave similar to a power set with complement, intersection and union. Use this tag for questions about Boolean algebras as structures, or about functions defined from/to Boolean algebras. For Boolean logic use the tag propositional-calculus.

Boolean algebras are structures which behave similar to a power set with complement, intersection and union. Use this tag for questions about Boolean algebras as structures, or about functions defined from/to Boolean algebras.

A Boolean algebra uses Boolean variables, typically denoted by capital letters, e.g. $A,B$, which can only take the values $0$ or $1$. Operators are $\land$ (conjunction), $\lor$ (disjunction) and $\lnot$ (negation).

For Boolean logic use the tag .

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How to prove this Boolean expression?

I'm trying to solve the following Boolean expression: A'B'C + A'BC' + A'BC + AB'C' + AB'C + ABC' + ABC = A + B + C I have tried using the distributive law and trying to cancel out some of the letters, and I keep getting something along the lines of…
bunkered_reageant
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Steps to simplify this boolean expression

How do you simplify: ~A*B+A*~C+B*~C to A * ~C + B * ~A I tried the distributive law but I end up going in circles.
James
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Proving a statement by contradiction

Not sure if many of you speak binary, but the question is to prove that 2n bits are sufficient to store the product of two unsigned n-bit numbers (i.e., there will be no overflow). I've thought of a simple method by contradiction, but its way too…
Snowman
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understanding Boolean algebra ( Boolean algebra simplification )

F = A'BC + AB'C + ABC' + ABC I know that by K-map answer is AB+AC+BC, and by boolean rules, it will be =A'BC + AB'C + ABC' + ABC+ ABC + ABC =(A'BC + ABC) + (AB'C + ABC) + (ABC' + ABC) =(A' + A) BC + (B' + B) CA + (C' + C) AB =AB+AC+BC I can't…
dana
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Circuit to Karnaugh map

A logical circuit was given, I had to create a boolean expression from it; which I did correctly (I guess). I calculated the output of each gate separately, considering the variables. From the given circuit I got this expression: ('xy + x) + ((x +…
O'Niel
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Rewriting a Boolean AND-OR as OR-AND

I have an expression a'b'c + a'bc + ab'c + abc' in AND-OR and I need to write as OR-AND. Simply applying De Morgans doesnt work as it results in: (a+b+c')(a+b'+c')(a'+b+c')(a'+b'+c) The truth table for this is all 1. Should I pick a value and use…
pstatix
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Prove that (a+bc) + (a'(b'+c')) = 1

Im trying to prove the boolean expression (a+bc) is the complement of (a'(b'+c')). To do that I need to prove that (a+bc) + (a'(b'+c')) = 1, however I only get as far as: (a + bc + a')(a + bc + b' + c') (a + a' + bc)(a + bc + b' + c') (1 + bc)(a +…
pstatix
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Proving the Boolean expression $(a'+b')((b'+c)+b'c) = b'+a'c$

Im trying to prove $(a'+b')((b'+c)+b'c) = b'+a'c$ and I am stuck on the second half. I know I need to get it to $(a'+b')(b'+c)$ to use the Distributive Property of OR over AND, but I cant seem to workout how to transform $((b'+c)+b'c)$ to $(b'+c)$.…
pstatix
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Boolean algebra proof - point me in the right direction?

I wish to formulate a proof that if $x+y = x+z$ and $xy$ = $xz$ then $y=z$. I'm just beginning my study of Boolean algebra, but is $y=z$ not self evident from the stated equations?
Jason Byrne
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Is it possible to apply De-Morgan's laws in the expression: $\rm [(A' + C) (A + B)']'$ and obtain $\rm (A'C)' + (AB)''$?

Is it possible to apply De-Morgan's laws in the expression: $\rm [(A' + C) (A + B)']'$ and obtain $\rm (A'C)' + (AB)''$?
Algorne
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Self dual functions

My book says that $f(x,y,z) = xy + yz + zx$ is self dual function but $(x+y)(y+z)(z+x)$ is not. I understood how $xy + yz + zx$ is self dual but I think $(x+y)(y+z)(z+x)$ is also self dual because in k-map if we represent $(x+y)(y+z)(z+x)$ then it's…
Zephyr
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Simplification of Karnaugh Table

\begin{array}{|c|c|c|c|} \hline abc/de& 000 & 001 & 011 & 010 & 110 & 111 & 101 & 100 \\ \hline 00& 1 & 0 & 0 & - & 1 & 0 & 1 & -\\ \hline 01& 0 & - & 1 & 0 & 0 & - & - & 0\\ \hline 11& 0 & - & 1 & 0 & 0 & - & - & 0 \\ \hline10 & 1 & - & 0 & - &…
roffensive
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Confused About Boolean Expression Simplification

I have this given problem: Expression Rule(s) Used 1. (A + C)(AD + AD') + AC + C Original Expression 2. (A + C)A(D + D') + AC + C Distributive. 3. (A + C)A + AC + C Complement, Identity. 4. A((A + C) + C) + C…
Phil
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how to prove bc + a'cd + ab'cd + bd' + bc'd = cd + b with Boolean algebra?

I'm trying to prove this equation using Boolean algebra : (it is a digital design problem) $$bc + \bar acd + a \bar bcd + b \bar d + b \bar cd = cd + b$$ What I'm done: $$bc + \bar acd + a \bar bcd + b \bar d + b \bar cd$$ $$=cd(\bar a + a \bar b) +…
Uncle
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Boolean expression help

I need to simplify the following, although can't find good examples to indicate how to achieve this. $$(A+\bar{B}).\bar{A}$$
James Thomas
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