Boolean Algebra

Canonical Form: Relation between min and max terms

The compliment of the function expressed in terms of sum of min terms can be obtained by taking sum of missing min terms in the original functions

e.g.  f= ∑(4, 5, 7) = m7 + m5 + m4  

Its compliment is              f’ = ∑(0, 1, 2, 3, 6) = m0 + m1 + m2 + m3 + m6

The compliment of the function expressed in terms of product of max terms can be obtained by taking product of max missing terms in the original functions

e.g. f= M4 M5 M2 M0 = π (0, 2, 4, 5)

Its compliment is              f’ = π (1, 3, 6, 7)= M7 M6 M3 M1

The compliment of the function expressed iin any of the canonical terms can be obtained by just interchanging the symbols π & ∑ and keeping the list of numbers same

e.g. f= ∑(4, 5, 7) = m7 + m5 + m4  

Its compliment is              f’ = π (4, 5, 7)= M4 M5 M7

e.g. f= M4 M5 M2 M0 = π (0, 2, 4, 5)

Its compliment is              f’∑(0, 2, 4, 5) = m0 + m2 + m4 + m5

The function expressed in terms of product of max terms can be converted to sum of min terms or vice-versa can be done by interchanging π & ∑ and list the numbers which were missing from the original function.

e.g. f= ∑(4, 5, 7) = m7 + m5 + m4  

then f can also be expressed as  f= ∑(4, 5, 7) = π(0, 1, 2, 3, 6)

e.g. f= π (0, 2, 4, 5)

then f can also be expressed as  f= π (0, 2, 4, 5)=∑(1, 3, 6, 7)

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