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Need to be sure of this boolean algebra problem's solution

  • Thread starter Amr719
  • Start date
31
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1. Homework Statement
Express the function Y= (abd + c)' + ((acd)'+(b)')' as the complete disjunctive normal form:
2.1 by applying Boole's theorerm,

2. Homework Equations


3. The Attempt at a Solution
I separated the equations to two terms (T1,T2)

T1= (abd + c)' T2=((acd)'+(b)')'

T1= (abd+c)' T2=((acd)'+(b)')'
=(abd)'.(c)' =(acd)".(b)''
=((a)'+(b)'+(c)'+(d)'). (c)' = abcd
= a'c'.(b+b') + b'c'(a+a') + c'd'(a+a')
=a'c'(bd+(bd)')+b'c'(ad+a'd')+c'd'(ab+a'b')
=a'c'bd+a'c'b'd'+b'c'ad+a'd'b'c'+abc'd'+a'b'c'd'


T1+T2= a'bc'd+ab'c'd+abc'd'+abcd
 

haruspex

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1= (abd+c)' T2=((acd)'+(b)')'
=(abd)'.(c)' =(acd)".(b)''
=((a)'+(b)'+(c)'+(d)'). (c)' = abcd
Not sure how to interpret your working. I think you intended this as two columns of working, the left hand for T1 and the right hand for T2. In which case, your last steps in what I quoted above are
=(abd)'.(c)' =((a)'+(b)'+(c)'+(d)')
And
=(acd)".(b)'' = abcd
Neither of those are correct. Take them in smaller steps.
 

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