boolean algebra simplify using kmap

Question

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 $ab+b'c$

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?

Answer 1

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:

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

$ab+b'c$