Boolean Help: Solving Your Problem with Ease

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The discussion revolves around simplifying Boolean expressions using XOR operations. Participants suggest various methods, including rearranging terms, applying properties of XOR, and generating truth tables. A key example demonstrates how to simplify the expression A'B XOR BC XOR AB XOR B'C' to ultimately arrive at C'. The conversation emphasizes the importance of understanding XOR properties, such as A XOR A' equating to 1, to achieve simplification. Overall, the thread provides insights into solving Boolean problems effectively.
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Can someone help me with this problem.
 

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i wanted to proof that that boolean is true. thank you

i wanted to proof that that boolean is true. thank you
 
I need to simplify the following expressions in XOR The simplified answer is only one variable or its complement
 
I can't see the JPEG yet. Why don't you explain the problem and what you've done so far?
 
Last edited:
One alternative - you can generate a truth table of the expression and see from the table what it is done, but I doubt that any Math teacher would like that, anyway

For example a
I believe that you can rearrange the terms and you can simplify them

e.g.

A'B XOR BC XOR AB XOR B'C'

becomes

A'B XOR AB XOR BC XOR B'C'

note:
A'B XOR AB - what does that tell you?

you can simplyfiy this to get just one variable

then take BC XOR B'C'

how can you get a more simple equivalent of this of a form Y XOR Z where Y and Z represent one variable e.g. B or C'.

When you have this done you should see the answer.

Example b is a tautology.
 
A'b Xor Ab = B?
 
Bc Xor B'c'
=(b Xor C)'
=b+c'
=b' Xor C = Bc + B'c'
= B=c?

So The Answer Would Be B Xor (b = C) = 0??
 
B Xor (b Ξ C)
 
  • #10
TO approve this other problem
AB XOR A'B XOR A'B' XOR B'A
B(A XOR A') XOR B'(A' XOR A)

we know that A XOR A' =1

B XOR B'

and from one of the XOR property, X XOR X' = 1
THEREFORE B EQUALY 1?
THE SOLUTION WOULD BE 1?
let me know if that right?
 
  • #11
i think i did the 1st problem wrong...
here is the simplied solution
A'B xor BC xor AB xor B'C'
becomes
A'B xor AB xor BC xor B'C'

A'B xor AB = B (we know that A A' =1)

BC xor B'C'
= (B xor C')(C xor B')

B xor (B xor C')(C xor B')
 
  • #12
A'B xor BC xor AB xor B'C'

this is just one possibility:

step 1. - rearrange (correct)

A'B xor AB xor BC xor B'C'

step 2 - A'B xor AB is equivalent to just using B (correct), we have now

B xor BC xor B'C'

step 3 - BC xor B'C' is equivalent to B xor C' we are allowed to do this, since substituiting this will not alter the value of the overall truth table

B xor B xor C'

step 4 - B xor B will always be 0 we have now

0 xor C'

step 5 - 0 xor C' depends only on the value of C'

result: C'

this is ofcorse just one possibility.

you got the B example right.

A'B xor AB = B (we know that A A' =1)
what has A A' got to do with it?
 
  • #13
Thank You So Much. That Really Helped.
Thanks Everyone
 
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