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Question Number 157725 by joki last updated on 27/Oct/21

form f(x,y,z) =((xy)′c)′((x′+c)(y′+z′))′   in standard SOP form and canonical SOP form

$$\mathrm{form}\:\mathrm{f}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\:=\left(\left(\mathrm{xy}\right)'\mathrm{c}\right)'\left(\left(\mathrm{x}'+\mathrm{c}\right)\left(\mathrm{y}'+\mathrm{z}'\right)\right)'\: \\ $$$$\mathrm{in}\:\mathrm{standard}\:\mathrm{SOP}\:\mathrm{form}\:\mathrm{and}\:\mathrm{canonical}\:\mathrm{SOP}\:\mathrm{form} \\ $$

Answered by Kunal12588 last updated on 27/Oct/21

[(xy)′c]′[(x′+c)(y′+z′)]′  =[((xy)′)′+c′][(x′+c)′+(y′+z′)′]  =(xy+c′)(xc′+yz)  =xyc′+xyz+xc′+yzc′  =xc′(y+1)+xyz+yzc′  SSOP=xc′+xyz+yzc′  CSOP=xc′(y+y′)(z+z′)+xyz(c+c′)+yzc′(x+x′)  =xyzc′+xyz′c′+xy′zc′+xy′z′c′+xyzc+x′yzc′

$$\left[\left({xy}\right)'{c}\right]'\left[\left({x}'+{c}\right)\left({y}'+{z}'\right)\right]' \\ $$$$=\left[\left(\left({xy}\right)'\right)'+{c}'\right]\left[\left({x}'+{c}\right)'+\left({y}'+{z}'\right)'\right] \\ $$$$=\left({xy}+{c}'\right)\left({xc}'+{yz}\right) \\ $$$$={xyc}'+{xyz}+{xc}'+{yzc}' \\ $$$$={xc}'\left({y}+\mathrm{1}\right)+{xyz}+{yzc}' \\ $$$${SSOP}={xc}'+{xyz}+{yzc}' \\ $$$${CSOP}={xc}'\left({y}+{y}'\right)\left({z}+{z}'\right)+{xyz}\left({c}+{c}'\right)+{yzc}'\left({x}+{x}'\right) \\ $$$$={xyzc}'+{xyz}'{c}'+{xy}'{zc}'+{xy}'{z}'{c}'+{xyzc}+{x}'{yzc}' \\ $$

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