Prove the equality ac abc ab c a 16 points by truth table
Prove the equality (a.c +a’).(b’+c) = a’.b’ + c
(a) (16 points) by truth table.
(b) (13 points) by Boolean algebra.
(c) (16 points) by proving its dual theorem using Boolean algebra.
Solution
(a) Truth table
(a.c+a\').(b\'+c) = a\'.b\' + c
(b) by Boolean algebra
(a.c +a\') .(b\'+c) = a\'.b\' + c
L.H.S. (a.c +a\') .(b\'+c)
= ab\'c +acc +a\'b\' +a\'c
= ab\'c +ac +a\'b\' + a\'c ------- (cc = c)
= ac(b\' + 1) + a\'b\' + a\'c
= ac + a\'b\' +a\'c ------------(b\'+1 = 1)
= a\'b\' + c(a+a\')
= a\'b\' + c -------(a+a\' = 1)
= a\'c\' +c = RHS
(c) proving dual theorem using Boolean algebra
(a.c + a\') . (b\' + c) = a\'.b\' + c
= ((a\' + c\') . a) + (bc\') = (a + b) .c\'
= (a\'.a + ac\') + bc\' = ac\' +bc\'
= 0 + ac\' +bc\' = ac\' + bc\'
= ac\' +bc\' = ac\' +bc\'
= (a+b)c\' = (a+b)c\'
Hence LHS = RHS
| a | b | c | a\' | a.c | a\'+a.c | b\' | b\'+c | (a.c+a\').(b\'+c) | a\'.b\' | a\'.b\'+c |
| 0 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 1 |
