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Boolean algebra for 5 var

  • Thread starter hime
  • Start date
  • #1
26
0

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.
 

Answers and Replies

  • #2
297
2
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.
 
  • #3
26
0
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
 
  • #4
297
2
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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