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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
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