How to simplify this boolean algebra (logic)

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Discussion Overview

The discussion revolves around simplifying a boolean algebra expression related to logic circuits. Participants explore various methods to achieve the simplest form of the expression h = (abc + abd + acd + bcd)'.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant begins by proposing to convert the expression into a product-of-sum form: h = (abc)'(abd)'(acd)'(bcd)'.
  • Another participant attempts to expand the expression further, providing a series of transformations but expresses uncertainty about the correctness of their approach.
  • A different participant introduces an example with a different boolean expression, suggesting a method of simplification that involves using identities like B + B' = 1.
  • Participants ask for clarification on how to add new terms into the expression during the simplification process.

Areas of Agreement / Disagreement

There is no consensus on the simplification process, as participants present different methods and express uncertainty about their correctness.

Contextual Notes

Participants do not clarify certain assumptions or steps in their transformations, leaving some mathematical processes unresolved.

aruwin
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This is related to logic circuits.
How do I simplify this boolean algebra(logic) into its simplest expression?

h = (abc + abd + acd + bcd)’

My first attempt would be to change it into product-of-sum.
h = (abc)'(abd)'(acd)'(bcd)'

and then what next?
 
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aruwin said:
This is related to logic circuits.
How do I simplify this boolean algebra(logic) into its simplest expression?

h = (abc + abd + acd + bcd)’

My first attempt would be to change it into product-of-sum.
h = (abc)'(abd)'(acd)'(bcd)'

and then what next?

h = (abc)'(abd)'(acd)'(bcd)'

= (a'+b'+c')(a'+b'+d')(a'+c'+d')(b'+c'+d')

= (a'+b'+c'+d)(a'+b'+c'+d')(a'+b'+c+d')(a'+b'+c'+d')(a'+b+c'+d')(a'+b'+c'+d')(a+b'+c'+d') (a'+b'+c'+d')

=(a'+b'+c'+d)(a'+b'+c+d')(a'+b+c'+d')(a+b'+c'+d')(a'+b'+c'+d')


I hope it is correct..
 
i_madini said:
h = (abc)'(abd)'(acd)'(bcd)'

= (a'+b'+c')(a'+b'+d')(a'+c'+d')(b'+c'+d')

= (a'+b'+c'+d)(a'+b'+c'+d')(a'+b'+c+d')(a'+b'+c'+d')(a'+b+c'+d')(a'+b'+c'+d')(a+b'+c'+d') (a'+b'+c'+d')

=(a'+b'+c'+d)(a'+b'+c+d')(a'+b+c'+d')(a+b'+c'+d')(a'+b'+c'+d')


I hope it is correct..

Could you explain to me how you add the new terms into the expression?
 
Let's assume we have :

F = A'+AB

= A'(B+B') + AB # B+B'= 1, will not affect the equation.

= A'B + (A'B' + AB) # Cancel each other.

= A'B

By the same way :)
 

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