Help understanding how two boolean equations were reduced

Thread Starter

jakk

Joined Nov 6, 2012
12
Ok, I had thought of trying that as well. If I do that, I get

Y=A'B'+A'BC'D'+AB'C'D'

I am having trouble finding the correct form for a consensus term, though.

If a consensus form for three terms is in the form

BC+B'D+CD

,then if I let B=A or B, I do not have a matching "CD" term.
 

WBahn

Joined Mar 31, 2012
26,398
Focus on the term (A'B' + A'BC'D') first. Is there a consensus term there that would let you simplify this?
 

Thread Starter

jakk

Joined Nov 6, 2012
12
Ok, let's try this again...

A'B'+A'BC'D'+AB'C'D'
A'(B'+BC'D')+AB'C'D'
A'(B'(C'D'+1)+BC'D')+AB'C'D'
A'(B'C'D'+B'+BC'D')+AB'C'D'
A'(C'D'(B+B')+B')+AB'C'D'
A'(C'D'+B')+AB'C'D'
A'C'D'+A'B'+AB'C'D'
A'C'D'+B'(A'+AC'D')
A'C'D'+B'(A'(C'D'+1)+AC'D')
A'C'D'+B'(A'C'D'+A'+AC'D')
A'C'D'+B'(A'+C'D'(A'+A))
A'C'D'+B'(A'+C'D')
A'C'D'+A'B'+B'C'D'
 
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