javar
Joined: 17 Apr 2008
Posts: 6
Helped: 1
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 12 May 2008 20:39 Re: proof of de morgan theorem |
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This is the proof:
1. (xy)'=x'+y'
x'+y'=(x'+y').1 Axiom(A.1=A)
= (x'+y')(xy+(xy)') Axiom (A+A'=1)
= x'xy+x'(xy)'+y'xy+y'(xy)'
the first and the third term are equal zero ( AA'B=(AA')B=0.B=0), then:
=x'(xy)'+y'(xy)'
=(xy)'(x'+y') Axiom distributive
=(xy)'(x'+y') + (xy)(xy)' Axiom(A+0=A)
=(xy)'(x'+y'+xy) Axiom distributive
=(xy)'(x'+y'+xy+xy) Axiom(A=A+A)
the first and third term in parenthesis as well as the second and
fourth are equivalent to x'+y+y'+x then
=(xy)'(x'+y+y'+x) Theorem (A+A'B=A+B)
=(xy)'(x'+x+y+y')
=(xy)'(1+1) Axiom (A+A'=1)
=(xy)'.1= (xy)' Axiom (A.1=A)
The theorem used in the above demonstration is easy to proof
A+A'B=(A+A')(A+B) Axiom distributive
=1.(A+B) Axiom(x+x'=1)
=A+B Axiom (x.1=x)
The second theorem is a demonstration Similar to previous simply exchanged at every step by operators OR per AND and 1 per 0 .
Discuss any doubt.
Good Luck!
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