All Questions
Tagged with propositional-calculus boolean-algebra
322
questions
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987
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Converting a Proposition to DNF using proof systems
I have been attempting to convent a prop to DNF using a group of common rules, i have applied them all but i think i should be able to get it smaller, This is what I've got so far. Thanks!
$$(p \wedge ...
4
votes
1
answer
4k
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Lindenbaum Algebras
After reading this page, I still have some questions about Lindenbaum algebras. Assume that the scope is a propositional language with a denumerable set X of propositonal variables.
In that case, the ...
- 41
3
votes
1
answer
139
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Chains in the Lindenbaum algebra
What is the easiest example of an infinite chain in a Lindenbaum algebra for the propositional calculus?
Does there exist an infinite antichain in a Lindenbaum algebra?
- 133
1
vote
1
answer
3k
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Expanding this boolean expression
Can this Boolean expression:
$$A*\overline{A*B}$$
be expanded to give:
$$A*\overline{A} * A*\overline{B}$$
Although that appears to reduce to zero?
I know $A(\overline{A+B})$ can be expanded to ...
- 1,525
3
votes
5
answers
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Algebraic proof of De Morgan's Theorems
Could someone give me an algebraic proof of De Morgan's Theorems?
I already know the graphic proof with the truth table, but I need to understand the algebraic way.
EDIT
I try to explain better. ...
- 483
0
votes
1
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398
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Simplify boolean expression
$(xy’+z)’\cdot((xz)’+y')$
$$\begin{align*}
(xy’+z)’\cdot ((xz)’+y’) &=(x'+yz’)\cdot (x’+z’+y’)\\
&=x’x’ + x’z’ + x’y’ + yz’x’ + yz’z’ + yz’y’\\
&=x’ + x’z’ + x’y’ + yz’x’ + yz’...
- 101
2
votes
2
answers
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Can $A+\bar{A}\bar{B}+BC$ get any simpler?
I've simplified this Boolean formula quite a bit. Can it get any simpler? My definition of simple in this case is using the least amount of operators (and, or)
Title is "A or (negative A and negative ...
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