How reduce this at simplest form using de Morgan's? #f=bar((barx+bary)+bar(barx+bary))#

2 Answers
Cesareo R.
Oct 15, 2017

#O/#

Explanation:

#f(x,y) = x y f(1,1)+x bar y f(1,0)+ bar x y f(0,1)+bar x bar y f(0,0)#

Here

#f(1,1) = bar((bar 1+bar 1)+bar((bar 1+bar 1))) = 0#
#f(1,0) = bar((bar 1+bar 0)+bar((bar 1+bar 0))) = 0#
#f(0,1) = bar((bar 0+bar 1)+bar((bar 0+bar 1))) = 0#
#f(0,0) = bar((bar 0+bar 0)+bar((bar 0+bar 0))) = 0#

then

#f(x,y) = O/#

Ratnaker Mehta
Oct 15, 2017

#0, or, phi.#

Explanation:

We will use the familiar rules of Boolean Algebra.

We have, #barx+bary=bar(x*y) rArr bar(barx+bary)=bar(bar(x*y))=x*y.#

Next, #barx+bary=bar(x*y).#

Combining these, we get,

#barx+bary+bar(barx+bary)=bar(x*y)+x*y=baru+u, say, where, "u=x*y,#

#"But, "u+baru=1.#

#:. barx+bary+bar(barx+bary)=1.#

#"Therefore, "bar{barx+bary+bar(barx+bary)}=bar1=0, or, phi.#

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