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Q.

Truth table for the system on four NAND gates as shown in figure is

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a

ABt000011101110

b

ABt000010101111

c

ABt001011100110

d

ABt001010100111

answer is A.

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Detailed Solution

γ=((A⋅B¯)⋅A¯)⋅((A⋅B¯)⋅B¯)¯Let A=0 & B=0                                           Let A = 1 & B = 0γ=((A⋅B¯)⋅A¯)⋅((A⋅B¯)⋅B¯)¯                    γ=((A⋅B¯)⋅A¯)⋅((A⋅B¯)⋅B¯)¯ =((0.0¯)⋅0¯).((0.0¯)⋅0¯)¯                                   =((1.0¯)⋅1¯)((1.0¯)⋅0¯)¯ =(0¯.0¯).(0¯.0¯)¯                                                =(0¯.1¯).(0¯.0¯)¯ =(1.0¯).(1.0¯)¯                                                =(1.1¯).(1.0¯)¯ =(0¯)⋅(0¯)¯                                                     =(1¯).(0¯)¯ =1.1¯                                                            =(0).(1)¯ =1¯                                                              =0¯ = 0                                                             = 1Let A=0 & B=1                                           Let A = 1 & B = 0γ=((A⋅B¯)⋅A¯)⋅((A⋅B¯)⋅B¯)¯                              γ=((A⋅B¯)⋅A¯)⋅((A⋅B¯)⋅B¯)¯ =(0.1¯.0¯).(0⋅1¯)⋅1¯)¯                                                 =(1.1¯.1¯).(1⋅1¯)⋅1¯)¯   =(0¯.0¯).(0¯.1)¯¯                                                       =(1¯.1¯).(1¯.1)¯¯    =(1.0¯).(1.1¯)¯                                                       =(0.1¯).(0.1¯)¯    =(0¯).(1¯)¯                                                             =(0¯).(0¯)¯ =(1).(0)¯                                                             =1.1¯ =0¯                                                                    =1¯ =1                                                                    =0  ABt000011                  option 1 is correct101110
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