Simplifying boolean expressions

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Discussion Overview

The discussion revolves around the simplification of a specific Boolean expression, (a+b+c')(a'b'+c), with participants attempting to reduce the number of literals involved. The focus is primarily on exploring different methods and approaches to achieve a simplified form, including the potential use of XOR operations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in simplifying the expression and reports no reduction in the number of literals after several attempts.
  • Another participant questions whether the use of XOR is permitted in the simplification process.
  • A different participant attempts to use XOR in their simplification, resulting in a new expression with one fewer literal but questions if this is the most simplified form.
  • Another participant begins a new transformation of the expression but does not complete their thought, indicating ongoing exploration.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method for simplification, and multiple approaches are presented without agreement on which is optimal. The discussion remains unresolved regarding the most simplified form of the expression.

Contextual Notes

Some participants express uncertainty about the use of XOR in this context, and there are indications of missing assumptions or steps in the simplification process that have not been fully explored.

Who May Find This Useful

This discussion may be of interest to students and practitioners of Boolean algebra, particularly those looking to understand various methods of simplification and the implications of using different operations like XOR.

Pi Face
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Homework Statement


Simplify the following Boolean expressions to a minimum number of literals
(a+b+c')(a'b'+c)

2. The attempt at a solution
Whenever I tried this I made no progress in reducing the number of literals, I just reordered the expression.

(a+b+c')(a'b'+c)
=aa'b'+a'bb'+a'b'c'+ac+bc+cc'
=0b'+a'0+a'b'c'+ac+bc+0
=0+0+a'+b'+c'+ac+bc+0
=a'b'c'+ac+bc
=a'b'c'+c(a+b)

I began with 6 literals and ended with 6. What else can I try?
 
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Pi Face said:

Homework Statement


Simplify the following Boolean expressions to a minimum number of literals
(a+b+c')(a'b'+c)

2. The attempt at a solution
Whenever I tried this I made no progress in reducing the number of literals, I just reordered the expression.

(a+b+c')(a'b'+c)
=aa'b'+a'bb'+a'b'c'+ac+bc+cc'
=0b'+a'0+a'b'c'+ac+bc+0
=0+0+a'+b'+c'+ac+bc+0
=a'b'c'+ac+bc
=a'b'c'+c(a+b)

I began with 6 literals and ended with 6. What else can I try?

Are you allowed to use XOR?
 
It doesn't say we can't and we did cover it, however it doesn't show up in any of the other problems or examples so far
 
I tried to use X(N)OR and this is what I came up with:

From
=a'b'c'+ac+bc

I can multiply ac and bc by (1) in the form (x+x')

=a'b'c'+ac(b+b')+bc(a+a')
=a'b'c'+abc+ab'c+abc+a'bc

Get rid of one of the two abc (redundant)

=a'b'c'+abc+ab'c+a'bc

Factor

=bc(a+a')+b'(ac+a'c')
=bc(1)+b'(aXORc)'
=bc+b'(aXORc)'

Now I have 5 literals, but is that really the most simplified term? I did all this work just to eliminate 1 literal :(
 
Pi Face said:
a'b'c'+ac+bc

= c'·a'b' + c(a+b)

= c'·a'b' + c·(a'b')'

= ...
 

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