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I've got the next operator: $L(W,X,Y)=(W+Y)X'$

I need to prove whether the operator is universal, And if it isn't is: {$L,1$} or {$L,0$} are universal.

I know that what i need to do is either showing that i can create and NAND or NOR table.

But i can't seem to understand how do i do that? Should i use boolean-table? And if so, How can i actually 'see' that i've created a NAND / NOR table with 3 parameters?

I'm pretty confused, And couldn't find a good example.

StationaryTraveller
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2 Answers2

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Hint: $L({1,X,1}) = x' => Inverter$

$L({W,0,Y})= W + Y => OR$

From OR and Inverter we can get a NOR gate

sai kiran grandhi
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You can get the constant 0:

L(x,x,x) = (x+x)x' = xx' = 0

and then:

L(W,L(W,W,W),Y) = L(W,0,Y) = W+Y

with using the constant 1:

L(1,L(W,L(W,W,W),Y),1) = (1+1)L(W,L(W,W,W),Y)' = L(W,L(W,W,W),Y)' = (W+Y)'

and you get a NOR operator with {L,1} only.

itai
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