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explain this output of CLC

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convert given f2 from expression to SOP form  we get F2= Sigma( 4 5  6 7  8  9 10 11 13 )

given : f1 = (1 4 5 8 10 12 14 15)

Since , F1 AND F2 = Common terms  i.e 4 5 8 10

f1 not given          
   

And gate  

         f1 and f2 =  commmon terms       
f2 4 5 6 7 8 9 10 11 13                                      
         1 4 5 8 10 12 14 15

OR gate 

  f=1 4 5 8 10 12 14 15
 

 

 

 

 

 

 

f3

To find

Not gate

 not(F3)= 1 12 14 15 + d(4 5 8 10 )      
             

Since , F1 AND F2 = Common terms  i.e 4 5 8 10

F3  is NOT gate so it will contain all elements which are not in F  : we get sigma ( 0 2 3  6 7 9 11 13  )  which is equivalent to pie( 1  12 14 15)

4 5 8 10 exits as dont care condition in f3 as it could be derived from f1 and f2 , i.e. it may not be present in F3

Hence answer is C!

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