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

AI Thread Summary
The discussion centers on whether the set B={0,1,R,F,X} qualifies as a Boolean algebra. Participants argue that the presence of five values complicates the application of Boolean principles, as true Boolean algebra is based on two-valued logic. Concerns are raised regarding the definitions of negation and conjunction for the elements, particularly with 'not X' and 'not F', which do not yield consistent results. The truth tables provided demonstrate contradictions, such as R and ~R not equating to 0 as required by Boolean axioms. Ultimately, the consensus suggests that B does not satisfy the necessary conditions to be classified as a 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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