In class I was asked to show how the expression in two of the Chapter 2 slides could be expanded into minterms (sum of products) and expanded into maxterms (product of sums).

F(A,B,C)	= AB + B'(A'+ C')

			  eliminate ( )'s

		= AB + A'B' + B'C'

			replace missing literals with (x+x')

		= AB(C+C') + A'B'(C+C') + (A+A')B'C'

					  eliminate ( )'s

		= ABC + ABC' + A'B'C + A'B'C' + AB'C' + A'B'C'
		
			replace with minterms, binary indices (x'-> 0)
			
		= m111 + m110 + m001 + m000 + m100 + m000
		
			replace binary indices with decimal indices
		
		= m7 + m6 + m1 + m0 + m4 + m0
		
			order and eliminate duplicates
		
		= m0 + m1 + m4 + m6 + m7 = sum m(0,1,4,6,7)

F is "1" when a minterm is one, or for inputs ABC = 000, 001, 010, 110, or 111
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To find maxterms (product of sums)

F(A,B,C)	= AB + B'(A'+ C')

eliminate ( )'s

		= AB + A'B' + B'C'

use the Distributivity rule, X+YZ = (X+Y)(X+Z)  let X -> AB + A'B'

			=  (AB + A'B' + B') (AB + A'B' + C')
			
			= (A + A'B' + B') (B + A'B' + B')  (A + A'B' + C') (B + A'B' + C')

use the Simplification  rule, X+X'Y = X+Y  to eliminate either A or A' from each sum term.

			= (A + B' + B') (B + A' + B')  (A + B' + C') (B + A' + C')
			
use XX' = 0, X + X' = 1,  X + X = X,   1 + X = 1, 1X = X
			
			= (A + B') ( 1 )  (A + B' + C') (A' + B + C')
			
use (X + Y) = (X + Y + ZZ') = (X + Y + Z) (X + Y + Z') to add missing literals to terms
					
			= (A + B' + C) (A + B' + C')  (A + B' + C') (A' + B + C')
		
F(A,B,C)	= (A + B' + C) (A + B' + C')  (A + B' + C') (A' + B + C')
		
Write as maxterms with binary indices (A' + B + C') -> 101:

F(A,B,C)	= M010  M011 M011 M101

			replace binary indices with decimal indices
		
		= M2 M3 M3 M5
		
			order and eliminate duplicates
			
F(A,B,C)	= M2 M3 M5 = Product M(2,3,5)

F is "0" when a maxterm is zero, or for inputs ABC = 010, 011, or 101

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Notice that the maxterm indices are just the set 0,1,2,3,4,5,6,7 less the minterm indices, and vice versa.

John Copeland