if f1 give the 2(0) since it is dont care and f2 gives(1) since its minterm then f will not contain the 2 as minterm in its function.

there is difference between dont care and minterm

minterm means value 1

dont care can be 1 and 0 .

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since f= f1.f2 so f will be equal to 1 iff f1=1 and f2 =1

f1 is 1 for the minterms 0,1,2,3,5,7 and f2 is 1 for the minterm 0,1,2,4,5,7

so f will be 1 for the minterm 0,1,2,5,7 where 7 is the dont care minterm .

so f = ∑m(1,5)+d(0,2,7)

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According to me F is AnD operation of f1 and f2 so in f minterm will be common min-term of both function.

since it is AND operation true is only for f1=1 f2=1.

and for dont care first take common and

then check the f1 dont care is present in minterm of f2 or not if yes then take it in(e.g. let 2 in f1 be 1 and 2 is in f2 so it will be 1) similarly check for f2 vice versa. this is done because . 1 and Dn't care is a don't care because the value will now depend on what i take the don't care to be . so we just take intersections of don't care with minterms of f2.

f1=∑m(0,1,5)+d(2,3,7)

f2=∑m(1,2,4,5)+d(0,7)

f=∑m(1,5)+d(0,2,7)

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