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1. The implication gate, shown below has two inputs ($x \text{ and }y)$; the output is 1 except when $x =1 \text{ and } y=0\text{, realize }f=\bar{x}y+x\bar{y}$ using only four implication gates.

2. Show that the implication gate is functionally complete.

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Is a figure here?
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There is no figure on the test paper I have. I have searched a few test papers and there is no figure on them.
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Okay.. Thanks :)
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There is a image.An or gate with bubbled input x.

@arjun this isn't functionally complete.The function x'+xy preserves 1 for f(1,1) so we can say that it is not functionally complete right?

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implication function is only partially complete

Good point.

Implication gate is A->B which becomes A'+B

So, let $f(A,B)=A'+B$

$f(A,0)=A'$ (we get complement )

$f(f(A,0),B)=f(A',B)=A+B$ (we get OR gate)

Thus it is functionally complete.

Let $F(X,Y) =X'+Y$

$F(Y,X)=Y'+X$

$F(F(Y'+X),0)=X'Y$

$F(F(X,Y),X'Y)=XY'+XY'$  Therefore, the above function is implemented with $4$ implication gates.
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implication is not fully functionally complete. Its partially functionally complete. Because it can derive NOT gate with the help of 0 as input.  I

Assuming the special block as representing  $(\bar{x} + y)$ with the bottom inverted, the $XOR\,(\bar{x}y + x\bar{y})$ expression can be derived as shown above using 4 implication gates.

But the implication function is only partially complete, i.e. can only represent a functionally complete set with an additional $0$ input. And that can be seen above where the gate in the middle represents a $NOT$ gate.