Verify the following using Boolean lows x’ + y’z = x’y’z’ + x’yz’ + x'yz + x’y’z + xy’z

 

Answer:

To prove:

x’ + y’z  = x’y’z’ + x’yz’ + x'yz + x’y’z + xy’z 

Proof:

RHS

= x’y’z’ + x’yz’ + x'yz + x’y’z + xy’z

= x’y’z + x’y’z’ + x’yz’ + x’yz + xy’z                                        (rearranging the terms)

= x’y’(z+z’) + x’y(z+z’) + xy’z                                                 (using distributive law)

= x’y’.1 + x’y.1 + xy’z                                                             (using complement law)

= x’y’ + x’y + xy’z                                                                   (using identity law)

= x’(y’+y) + xy’z                                                                     (using distributive law)

= x’.1 + xy’z                                                                           (using complement law)

= x’ + xy’z                                                                              (using identity law)

= (x+x’).(x’+y’z)                                                                     (using distributive law)

= 1.(x’+ y’z)                                                                           (using complement law)

= x’ + y’z                                                                                (using identity law)

= LHS

Hence, the expression x’ + y’z  = x’y’z’ + x’yz’ + x'yz + x’y’z + xy’z is verified.

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.