Concept of Dual!?

[Steven O.]

I'm just learning digital electronics, taking an introductory class.
Everything is pretty clear so far, except the concept of a "dual" of a
logical function has me slightly puzzled.

I think it's like this: If I take any true logical equation, and
reverse the operators, and interchange 1s and 0s, (and, do NOT switch
A to A' or vice-versa), the equation I get as a result is still true,
and is the dual of the original -- but the new equation is NOT the
equivalent of the orginal. Is that right?

For example:

A + A' = 1
AA' = 0

Or another example:

A + 1 = 1
A0 = 0

Or, once more:

A + 0 = A
A1 = A

Are each of these pairs, in fact, the dual of each other? Thanks in
advance for all replies.

Steve O.


"Spying On The College Of Your Choice" -- How to pick the college that is the Best Match for a high school student's needs.
www.SpyingOnTheCollegeOfYourChoice.com

 

[Ratch]

Duality: Every Boolean expression remains valid if the operations and
identity elements are interchanged.

Ex: If (x+y)' = x' * y' holds, then (x * y)' = x' + y' also holds.

Ex: If x + 1 = 1 holds, then x * 0 = 0

Ratch

I'm just learning digital electronics, taking an introductory class.
Everything is pretty clear so far, except the concept of a "dual" of a
logical function has me slightly puzzled.

I think it's like this: If I take any true logical equation, and
reverse the operators, and interchange 1s and 0s, (and, do NOT switch
A to A' or vice-versa), the equation I get as a result is still true,
and is the dual of the original -- but the new equation is NOT the
equivalent of the orginal. Is that right?

For example:

A + A' = 1
AA' = 0

Or another example:

A + 1 = 1
A0 = 0

Or, once more:

A + 0 = A
A1 = A

Are each of these pairs, in fact, the dual of each other? Thanks in
advance for all replies.

Steve O.


"Spying On The College Of Your Choice" -- How to pick the college that is

the Best Match for a high school student's needs.

 

[William Elliot]

Duality: Every Boolean expression remains valid if the operations and
identity elements are interchanged.

Ex: If (x+y)' = x' * y' holds, then (x * y)' = x' + y' also holds.
Ex: If x + 1 = 1 holds, then x * 0 = 0

To prove AA' = 0 from A + A' = 1 use DeMorgan
(A + A')' = 1'
A' A" = 0
A' A = 0
AA' = 0
(A + 1)' = 1'
A' 1' = 0
A' 0 = 0

but as this is for all A, we have for all A
(A')' 0 = 0
A0 = 0
(A + 0)' = A'
A' 0' = A'
A' 1 = A'

again as this is for all A,
A" 1 = A"
A1 = A

Yes, and now you know the potency of DeMorgan's rules and how
to derive the dual of any universal equation.

 

[Michael Barr]

I'm just learning digital electronics, taking an introductory class.
Everything is pretty clear so far, except the concept of a "dual" of a
logical function has me slightly puzzled.

I think it's like this: If I take any true logical equation, and
reverse the operators, and interchange 1s and 0s, (and, do NOT switch
A to A' or vice-versa), the equation I get as a result is still true,
and is the dual of the original -- but the new equation is NOT the
equivalent of the orginal. Is that right?

For example:

A + A' = 1
AA' = 0

Or another example:

A + 1 = 1
A0 = 0

Or, once more:

A + 0 = A
A1 = A

Are each of these pairs, in fact, the dual of each other? Thanks in
advance for all replies.

Steve O.


"Spying On The College Of Your Choice" -- How to pick the college that is the Best Match for a high school student's needs.
www.SpyingOnTheCollegeOfYourChoice.com

These are NOT (with one exception) the dual equations, although they
are all valid. The dual expression is gotten by also complementing
each variable. So the dual of A + 0 = A is A'1 = A'. The dual of A +
B = C is A'B' = C'. Note that the latter equation is not a tautology,
which all of your examples were. Now it is a characteristic of a
tautology that if you replace each free variable by an arbitrary
expression, you still get a tautology. So if you begin with A + 0 = A
and replace A by A', you get the tautology A' + 0 = A'. If you now
dualize, you get A1 = A, again a tautology. Now A + B = C is a
contingent expression; its truth depends on those of A, B and C, but A
+ B = C for exactly the same values as A'B' = C' and NOT those that
make AB = C true.

So to repeat, to dualize an expression (or an equation), exchange 0
and 1, exchange meet and join and complement each variable.

 

[Steven O.]

Thank you.

Steve O.

These are NOT (with one exception) the dual equations, although they
are all valid. The dual expression is gotten by also complementing
each variable. So the dual of A + 0 = A is A'1 = A'. The dual of A +
B = C is A'B' = C'. Note that the latter equation is not a tautology,
which all of your examples were. Now it is a characteristic of a
tautology that if you replace each free variable by an arbitrary
expression, you still get a tautology. So if you begin with A + 0 = A
and replace A by A', you get the tautology A' + 0 = A'. If you now
dualize, you get A1 = A, again a tautology. Now A + B = C is a
contingent expression; its truth depends on those of A, B and C, but A
+ B = C for exactly the same values as A'B' = C' and NOT those that
make AB = C true.

So to repeat, to dualize an expression (or an equation), exchange 0
and 1, exchange meet and join and complement each variable.

"Spying On The College Of Your Choice" -- How to pick the college that is the Best Match for a high school student's needs.
www.SpyingOnTheCollegeOfYourChoice.com