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Solve this. 

closed with the note: Got the answer
in Digital Logic by (181 points)
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f = D0 + D1
  = (~A1 . ~A0) + (~A1 . A0)
  = (~A1)
  = ~(D2 . D3)
  = ~( (A1 . ~A0) . (A1 . A0) )
  = ~( (x . ~y) . (x . y) )
  = (~(x . ~y) + ~(x.y) )
  = ( (~x + y) + ~x + ~y  )  = (~x + ~x + y + ~y) = 1
 option d)
0
1 is the answer

1 Answer

+3 votes
Best answer
Input to the 2nd Decoder is x'y.xy = 0

f = A0'A1' + A0A1' = z' + z = 1

Thus the Answer must be 1 .

Correct me if I went wrong somewhere.
by Loyal (5.8k points)
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