Is B={0,1,R,F,X} a Boolean Algebra?

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Discussion Overview

The discussion centers on whether the set B={0,1,R,F,X} constitutes a Boolean algebra. Participants are examining this question through the lens of basic postulates and axioms of Boolean algebra, including the creation of truth tables and logical operations.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that B is a Boolean algebra because it is possible to create AND, OR, and NOT tables with the elements of B.
  • Another participant challenges this assertion by questioning the logical consistency of operations, particularly the negation of elements like 'X' and 'F', suggesting that they cannot be definitively assigned a single value.
  • A truth table is presented by a participant, showing the results of AND and NOT operations for the elements in B, which indicates that R' equals F and vice versa.
  • Another participant points out a contradiction in the truth tables, noting that the identity A and ~A = 0 should hold, but the provided tables suggest that R and ~R yield X instead of 0, raising concerns about the validity of B as a Boolean algebra.

Areas of Agreement / Disagreement

Participants do not reach a consensus. There are competing views regarding the validity of B as a Boolean algebra, with some arguing in favor and others presenting logical challenges that suggest it may not satisfy the necessary conditions.

Contextual Notes

Limitations include the ambiguity surrounding the definitions and interpretations of the elements R, F, and X, as well as the implications of their interactions in the context of Boolean operations.

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Homework Statement



is B={0,1,R,F,X) a boolean algebra? Use basic posulates/axioms to prove it.

R=Rising
F=Falling
X=Dont Care

Homework Equations


Reference: Boolean Identities Table


The Attempt at a Solution


it is boolean algebra. you can create and, or , not tables with it.

please help.
 
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Did you try and make some logic tables for it? Doesn't seem so.

You'd end up with problems if you did. For example, what is 'not X'?

Or what is 'not F'? Surely it can't be R since if it's not falling, it could be steady. So not necessarily R, logically speaking.

What is '0 and R'?

I'm not sure what your prof is expecting here, exactly, but perhaps that will help. Would be good to know what axioms and postulates he has introduced.

To my mind, since boolean algebra could be described as "two-valued logic", it's obviously not true with 5 values that cannot be reduced to be equivalent to 2 values.
 
AND | 0 1 R F X
---- |---------
---0 | 0 0 0 0 0
---1 | 0 1 R F X
---R | 0 R R X X
---F | 0 F X F X
---X | 0 X X X X

NOT
x |~x
------
0 |1
1 |0
R |F
F |R
X |X

so obviously R' =F and vice versa
 
Well, let's see. We have one identity that is:

A and ~A = 0

So we should get:

R and ~R = 0

However, by your truth tables, ~R = F, that means:

R and ~R = R and F = X

However, it must equal 0, not X.
 

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