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question: this is standar sop,try to simplify it using kmap

           ----------  --------- 
        AB\CD    00    01    11   10 
         00       0     0    1     1
         01       0     0    0     0  
         11       1     1    1     1  
         10       0     0    1     1  

this is the right kmap. the answer is $ab+b'c$ enter image description here

enter image description here

this is the wrong one. suppose I did it like this. can I simplify this to get the right kmap?

=$ab+ac+a'b'c$
=$c.!(a+a'b')+ab$ -> applied demorgan, am I right?
=$c(a'.(a+b))+ab$
=$c(a'a+a'b)+ab$
=$a'bc+ab$
=$b(a'c+a)$ ->should I apply demorgan again? , but after I apply it, I didn't get =$ab+b'c$

is my assumption true?
Boolean algebra :

$AB+A'B+AB'$
=A(B+B')+B(A+A') -> why $A'B$ changed to B(A+A') , what law is this?

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Your mistake is in the very first step:

=$ab+ac+a'b'c$
=$c.!(a+a'b')+ab$ -> applied demorgan, am I right?

No, this is not right. When you take out the $c$, you simply get:

$ab+ac+a'b'c$
=$c.(a+a'b')+ab$

Anyway, I would show the equivalence differently: if you look at the K-Map you can see how the second expression can be put into the first:

First, break up the $ac$ group:

$ab+ac+a'b'c = \text{ (Adjacency)}$

$ab+abc+ab'c+a'b'c$

Then, the $ab$ term absorbs the $abc$ term:

$ab+abc+ab'c+a'b'c = \text{ (Absorption)}$

$ab+ab'c+a'b'c$

Finally, combine the last two terms into the $b'c$ term:

$ab+ab'c+a'b'c = \text{ (Adjacency)}$

$ab+b'c$

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  • $\begingroup$ Thankyou so much!!!! And also is there any way to know that the kmap result is not the most simplify one? Or like should i do trial and error to find the most simplify one? And can i use demorgan theorem for this? $\endgroup$ – fiksx Dec 14 '17 at 6:49
  • $\begingroup$ @Vixf Well, the general thing to do is to look for as few groups as ppossoble, each of which is as large as possible .... just looking at the kmap will typically tell you whether you did that, though when you start adding more variables it becomes harder and harder to see. And yes, you could do DeMorgan on something like $a'b'c'$ and turn that into $(a+b+c)'$, but that is typically not regarded as 'simpler' ... oftentimes the goal is to get an expression that's in CNF or DNF. $\endgroup$ – Bram28 Dec 14 '17 at 14:05

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