How many 2-input and gates can replace n-input and gate Does that apply for nand gates as well?

Answer 1

for a two input gate to represent as an n-input gate excatly n-1 two input gates are required. this implies that for a two input OR gate to represent a four input OR gate exactly three two input OR gates are required

let F is =a+b+c+d

=(((a+b)+c)+d)

=((a+b)+(c+d))

in both the above cases + is used three times so three two input OR gates make a four input OR gates.

This discussion doesnot hold good for NAND gates an example can illlustrate the reson:-

take F=(a.b.c.d)'=a'+b'+c'+d' --------------------------->(1)

(this is obtained by a four input NAND gate)

let us take this in the manner we did it for an OR gate and we will then verify the result.

=((a.b)'(c.d)')'

=((a'+b').(c'+d'))'

=(a'+b')'+(c'+d')'

=ab+cd <------------------------(2)

(1)is not equal to (2) so we can say that a NAND gate cannot be replaced in the manner as OR gate is replaced