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Forging new generations of engineers
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DeMorgan’s Theorems
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DeMorgan’s Theorem #1
A · B = A + B
EQUAL
A B A x B A x B A B A + B
0 0 0 1 0 0 1
0 1 0 1 0 1 1
1 0 0 1 1 0 1
1 1 1 0 1 1 0
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DeMorgan’s Theorem #2
A B A + B A + B A B A x B
0 0 0 1 0 0 1
0 1 1 0 0 1 0
1 0 1 0 1 0 0
1 1 1 0 1 1 0
A + B = A · B
EQUAL
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DeMorgan’s Theorem #1
A · B = A + B
B
A
B
A
=
Invert output of an
AND gate
Invert the inputs of an
OR gate
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DeMorgan’s Theorem #2
A + B = A · B
=
Invert output of an
OR gate
Invert the inputs of an
AND gate
B
A
B
A
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DeMorganizing an Expression
XCBCACBCA =))(+(+)(+)(
DeMorg. XCBCACBCA =)))(+(())((
DeMorg. XCBCACBCA =)))(+()(+)((
Dist. XCBCACACBCA =)))(+()(+(
XCBCACBCA =)))(+()()((AA =
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DeMorganizing an Expression
XCBCACACBCA =)))(+()(+(
0=AA XCBCAABCA =)))(+()(0+(
0=0A XCBCABCA =)))(+()(0+(
DeMorg
XCBCABCA =))(+)+)(((
XCBCABCA =)))(+()((AA =0+
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DeMorganizing an Expression
XCBCABCA =))(+)+)(((
DeMorg XCBCABCA =))+(+))(((
Assoc. XCBCABCA =)++)((
Dist. XCBCABBCACABCA =++
0=0A XBABBCACBC =0++0
XBBCA =0++0AA =0+
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DeMorganizing an Expression
XBBCA =0++0
AA =0+ XBBCA =
XBCA =AAA =

De Morgan Theorem B[1]