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

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SUMMARY

The discussion confirms that the set B={0,1,R,F,X} does not constitute a Boolean algebra due to its five distinct values, which cannot be reduced to the two-valued logic fundamental to Boolean algebra. Participants analyzed the implications of logical operations such as AND, OR, and NOT on the elements of set B, revealing inconsistencies, particularly with the operation of NOT on R and F. The inability to satisfy the identity A and ~A = 0 further substantiates that B fails to meet the axioms of Boolean algebra.

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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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