Disjunctive Normal form to Combinatorial Circuit

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The discussion revolves around converting a disjunctive normal form (DNF) into a combinatorial circuit. The provided DNF, f(x,y,z)=xyz+Not(xy)z, is identified as incorrect due to misunderstandings about the terms involved. Specifically, the term Not(xy) is clarified as not being equivalent to Not(x)Not(y), which is crucial for accurate representation. The participant realizes their mistake regarding the row analysis and acknowledges the need for proper negation in the terms. Ultimately, the expression cannot be simplified or factored further.
Kingyou123
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Homework Statement


Draw the combinatorial circuit corresponding to the disjunctive normal form.
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Homework Equations


DNF of f(x,y,z)=xyz+Not(xy)z

The Attempt at a Solution


f(x,y,z)=xyz+not(xy)z
=z(xy+not(xy))
Wouldn't the xy and not xy cancel out? That's my current problem with this probelm.
 
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The problem is that your DNF is not correct. Presumably the first term is supposed to give the 1 in the second row of the last column. But z is 0 in that row. So why are none of x,y,z negated in the term?
Your second term is not correct either. Are you aware that not(xy) is not the same as not(x)not(y)? de Morgan's Laws can sort that out for you.
 
andrewkirk said:
The problem is that your DNF is not correct. Presumably the first term is supposed to give the 1 in the second row of the last column. But z is 0 in that row. So why are none of x,y,z negated in the term?
Your second term is not correct either. Are you aware that not(xy) is not the same as not(x)not(y)? de Morgan's Laws can sort that out for you.
Oh thank you for catching my mistake, I was looking at row 0 for some reason.
So I'm left with xynot(z) + not(x)not(y)z, wouldn't everything just cancel out?
 
Kingyou123 said:
wouldn't everything just cancel out?
No. In fact the expression cannot be factorized at all.
 

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