But it is partially functionally complete.

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Since Not is not a binary function i think it is not required to derive it.

If external 1 is not supplied then also we can get A+B by using two half adders where the inputs of the 2nd half adder will be the Sum S_{1} and the Carry C_{1} of the 1st half adder.

So S_{2}=S_{1}⊕ C_{1}=(A ⊕ B) ⊕ (AB) = A+B.

(A ⊕ B) ⊕ (AB)

=(AB'+A'B)(AB)' + (A'B'+AB)(AB)

= (AB'+A'B)(A'+B')+ (0+AB)

=(0+A'B+AB'+0)+AB

= A'B+AB'+AB

= B(A'+A)+AB'

=B+AB'

=A+B

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