Boolean Help: Solving Your Problem with Ease

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Discussion Overview

The discussion revolves around solving and simplifying Boolean expressions, particularly focusing on the XOR operation. Participants seek assistance with specific problems and share various methods for simplification, including truth tables and properties of XOR.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants express a need for help with proving the truth of a Boolean expression.
  • One participant suggests using the definition of XOR in terms of AND and OR, while others propose working directly with XOR.
  • A participant mentions generating a truth table as a method for simplification, although they express doubt about its acceptance in academic settings.
  • Another participant provides an example of simplifying an expression involving multiple XOR operations, suggesting that certain terms can be rearranged and simplified.
  • There are discussions about specific expressions, with participants questioning the equivalence of certain simplified forms and exploring properties of XOR.
  • One participant claims that a certain expression simplifies to 1 based on known properties of XOR.
  • Another participant revisits their earlier work, attempting to clarify their simplification steps and the relevance of certain Boolean identities.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of specific simplifications or the final outcomes of the expressions discussed. Multiple competing views and methods for simplification are presented, and some participants express uncertainty about their results.

Contextual Notes

Some participants' claims depend on specific properties of Boolean algebra, and there are unresolved steps in the simplification processes. The discussion includes various assumptions about the expressions and their transformations.

Who May Find This Useful

This discussion may be useful for individuals interested in Boolean algebra, particularly those looking for different methods of simplification and understanding of XOR properties.

krispiekr3am
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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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