Duality and Demorgan's law

Question

Hi, What is the difference between "Dulaity" and "Demorgan's law"...Please state some examples.. so that my confusion gets resolved.

Answer 1

DUAL: if  you change

           (1) OR <-----> AND

             (2) 0<-------->1 

Demorgan's law :         

           (1)OR <-----> AND

            (2) 0<-------->1          
            (3)  literal's is complimented

ie dual property is subset of Demorgan's property.

example

let expⁿ  is   v + ∼u (x+y∼z)

(∼ means NOT operator)

dual : v.(∼u+(x.(y+∼z)))

for Demorgan's just the compliment the literals of DUAL expⁿ

 Demorgan's : ∼v.(u+(∼x.(∼y+z)))

Comments

it says that dual of f1 is f2....but vice versa is not true...

 

f2 is abc+ab+bc+ac  but in this answer to find out its dual  they are complementing literals also...which is not in defination of dual

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f₁=Σ(0,1,2,4)

f1=(∼a∼b∼c ) + (∼a∼bc) +(∼ab∼c ) + (a∼b∼c)

dual of f1 =(∼a+∼b+∼c ) .(∼a+∼b+c) .(∼a+b+∼c ) + (a+∼b+∼c) 

∏M(3,5,6,7)       (in maxterm)

=Σm(0,1,2,4) (in minterms)

so f1 is self dual

f₂=abc+ab+bc+ac

=Σm(3,5,6,7)

dual of f2 = (a+b+c)(a+b)(b+c)(a+c)

∏M(0,1,2,4)

Σm(3,5,6,7)

so f2 is also self dual

also dual of each other

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on dual literal can't complimented

see these problem

https://gateoverflow.in/751/gate2001-10

https://gateoverflow.in/100979/ace-self-dual