2019-11-11T17:33:47-05:00
Use properties of Boolean algebra to simplify the following Boolean expression (show-
ing all the steps):
(x + y
0
)(x
0 + y
0
)
0
1
2019-11-13T11:55:58-0500
( x + y ‾ ) ( x ‾ + y ‾ ) ‾ ( x + \overline y) \overline{(\overline x +\overline y)} ( x + y ) ( x + y ) we know,
A + B ‾ = A ‾ . B ‾ a n d A ‾ ‾ = A \overline {A+ B} = \overline A . \overline B \space and \space \overline{\overline A} = A A + B = A . B an d A = A
( x + y ‾ ) ( x ‾ + y ‾ ) ‾ = ( x + y ‾ ) ( ( x ‾ ‾ ∗ y ‾ ‾ ) ) ( x + \overline y) \overline{(\overline x +\overline y)} = ( x + \overline y) ( \overline{(\overline x} * \overline{\overline y})) ( x + y ) ( x + y ) = ( x + y ) ( ( x ∗ y ))
= ( x + y ‾ ) ( x ∗ y ) = (x + \overline y) ( x *y) = ( x + y ) ( x ∗ y )
= x . x . y + x y . y ‾ = x . y + 0 = x . y = x . x . y + x y .\overline y = x . y + 0 = x. y = x . x . y + x y . y = x . y + 0 = x . y Since
( x . x = x a n d y . y ‾ = 0 ) ( x . x = x \space and \space y .\overline y = 0) ( x . x = x an d y . y = 0 ) Answer :
( x + y ‾ ) ( x ‾ + y ‾ ) ‾ = x . y ( x + \overline y) \overline{(\overline x +\overline y)} = x. y ( x + y ) ( x + y ) = x . y
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