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

(y' + x) ∙ (z + z') ∙ (y' + x') + (z + x'∙z) ∙ (x + y)

This is what I have so far.

(y' + x) ∙ (z + z') ∙ (y' + x') + (z + x'∙z) ∙ (x + y)

= (y' + x) ∙ 1 ∙ (y' + x') + (z + x'∙z) ∙ (x + y) (Using Complement Element, x+x' = 1)

= (y' + x) ∙ (y' + x') + (z + x'∙z) ∙ (x + y) (Using Identity Element, 1 . x = x . 1 = x)

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?

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Here’s a start:

$$\begin{align*} (y'+x)\cdot(y'+x')&=y'\cdot(y'+x')+x\cdot(y'+x')\\ &=y'\cdot y'+y'\cdot x'+x\cdot y'+x\cdot x'\\ &=y'+x'\cdot y'+x\cdot y'\\ &=y'+(x'+x)\cdot y'\\ &=y'+y'\\ &=y' \end{align*}$$

Note also that $z+x'\cdot z=(1+x')\cdot z=z$.

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  • $\begingroup$ Thanks for the helpful reply! :) I shall go and try now. :) $\endgroup$ – Lawrence Wong Feb 3 '13 at 6:36
  • $\begingroup$ @Lawrence: You’re welcome; good luck with it! $\endgroup$ – Brian M. Scott Feb 3 '13 at 6:37
  • $\begingroup$ Hi again, I worked out the whole thing and I got y' + z * (x+y) Can it be simplify further? $\endgroup$ – Lawrence Wong Feb 3 '13 at 6:53
  • $\begingroup$ @Lawrence: Not really, though $y'+x\cdot z+y\cdot z$ might be considered simpler in many contexts, because it’s in disjunctive normal form. $\endgroup$ – Brian M. Scott Feb 3 '13 at 6:55
  • $\begingroup$ Hi again, I need help with a new question. Please take a look. :) $\endgroup$ – Lawrence Wong Feb 3 '13 at 7:32

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