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WorksheetsBoolean/DeMorgan's Theorems
Total questions: 22
Worksheet time: 2hrs 50mins
Simplifying the expression F = X̅Y + Y̅Z̅ + XY + Y̅Z yields:
F = X + Y̅
F = X + Y
F = 0
F = 1
______ ______
(A + B) (A̅+ B̅)
simplifies to
A
B
0
1
The Boolean expression (X + Y)(X + Y̅) simplifies to
X
Y
XY
X̅Y̅
Simplify the Boolean function x(x̅+y)
xx̅
xy
xy̅
x̅y̅
Simplify the Boolean function x + (x̅y)
xy
x̅ y̅
x+y
0
Complete the theorem: X ⋅ X =
0
1
X
__
X
Complete the theorem: X + X =
0
1
X
__
X
Complete the theorem: X + 0 =
X
__
X
0
1
__
X ⋅ X =
0
1
X
__
X
A ⋅ B
is the same as
B ⋅ A
A ⋅ A
B ⋅ B
A + B
__ __
A ⋅ B
is the same as
_______
A + B
A ⋅ B
B ⋅ A
______
A ⋅ B
X ⋅ ( Y ⋅ Z )
is the same as
X ⋅ Y + Z
( X + Y ) + Z
( X ⋅ Y ) ⋅ Z
( X ⋅ Z ) + Y
A ⋅ ( B + C )
is the same as
___________
A + B + C
( A + B ) ( A + C)
A ⋅ B ⋅ C
( A ⋅ B ) + ( A ⋅ C )
____
A B
is the same as
__ __
A + B
__ __
A B
( A B )
( A + B )
_________
(A (A+C))
is equivalent to ______
A
C̅
A̅
1
Simplify the expression (A+B)(A+C) = ____
A+C
B+C
A+BC
A
The Boolean expression A+AB simplifies to
A
B
A+B
1
(Q+Q)(RS̅+QR̅S̅) =
R̅
QS̅
1
(Q+Q)(RS̅+QR̅S̅)
MP̅+MP =
1
MP
M
MP̅+MP
(LM+N̅+S̅)(KK̅) =
0
KLM+S̅
(LM+N̅+S̅)(KK̅)
K̅KLM+KK̅N̅+KK̅S̅
(GH+G̅)G =
0
G
GH
(GH+G̅)G
(WZ1+0)W̅+1 =
1
Z
WZ
(WZ1+0)W+1
