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Show that $\{1,A \bar{B}\}$ is functionality complete, i.e., any Boolean function with variables $A$ and $B$ can be expressed using these two primitives.

We have $1$ and $AB'$.

$F(A,B) = AB'$

$F(1,B) = 1.B' = B.$

$F(A,B') = A.(B')' = AB$

We can $[*,’]$ so its functionally complete.
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