Simplifying the expression using Boolean Algebra Part 2

Question

Simplifying the expression using Boolean Algebra into sum-of-products (SOP) expressions . refers to AND + refers to OR

Updated with new question

( (a + b) ∙ (a' + c') )' + (b + c')' + a∙b'∙c

= ( (a + b) ∙ (a' + c') )' + (b' . c) + a∙b'∙c

= (( a ∙ (a' + c') + b ∙ (a' + c') )' + (b' . c) + a∙b'∙c

= (( a ∙ a' + a ∙ c') + (a' . b + b ∙ c') )' + (b' . c) + a∙b'∙c

= (( a ∙ c') + (a' . b + b ∙ c') )' + (b' . c) + a∙b'∙c

= ( a ∙ c')' . (a'. b + b ∙ c')' + (b' . c) + a∙b'∙c

= ( a' + c) . (a + b' ∙ b' + c) + (b' . c) + a∙b'∙c

I am stuck here. How do I continue? Any help please?

Answer 1

I’m going to omit the dots and just write $ab'$, for instance, instead of $a\cdot b'$. I think that your approach is reasonable down to here:

$$\big((ac')+(a'b + bc')\big)'+(b'c)+ab'c\;.$$

My next step would be to combine the last two terms: $b'c+ab'c=(1+a)b'c=b'c$. Now let’s worry about the first term. I think that I might do a little more manipulation inside before getting rid of the outer negation:

$$(ac')+(a'b+bc')=ac'+b(a'+c')\;,$$

so

$$\begin{align*} \big((ac')+(a'b + bc')\big)'&=\big(ac'+b(a'+c')\big)'\\ &=(ac')'\big(b(a'+c')\big)'\\ &=(a'+c)\big(b'+(a'+c')'\big)\\ &=(a'+c)(b'+ac)\;. \end{align*}$$

See what you can do from there.

However, you might be able to do better with a different start. Note that the original first term is $\big((a+b)(a'+c')\big)'$, which when expanded is going to have the very nice expression $(a'+c')'=ac$. Thus, we might want to start with

$$\begin{align*} \big((a+b)(a'+c')\big)'+(b+c')'+ab'c&=(a+b)'+ac+b'c+ab'c\\ &=a'b'+ac+b'c\;. \end{align*}$$

This can be manipulated in various ways, but none is an improvement.

Answer 2

Let us simplify the first term: $$[(a+b).(a'+c')]'$$ $$=(a+b)'+(a'+c')'$$ $$=a'.b'+a.c$$

So the required simplified form is $$a'.b'+a.c+a.b'.c$$ $$=a'.b'+a.c.(1+b')$$ $$=a'.b'+a.c$$

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Edit: I forgot the second term $(b+c')'$. So adding it, the final answer is $$a'.b'+b'.c+a.c$$