Detailed Video Solution: https://www.youtube.com/watch?v=XmmkE4yRk5c&t=3795s
$S = \{ 1, f(A,B) = A \bar{B} \}$ is functionally complete.
Proof 1: By creating $\{NOT, AND \}$ from $S:$
Creating "NOT": $f(1,B) = \bar{B} $ ; So we can create "NOT" operation using $S.$
Creating "AND": $f( A, f(1,B) ) = AB$ ; So we can create "AND" operation using $S.$
Since we create $\{NOT, AND \}$ from $S$, So, $S$ is functionally complete.
Proof 2: Using Post's functional completeness theorem:
$S = \{ 1, A \bar{B} \}$
1. "$1$" is Not 0-preserving.
2. $A \bar{B} $ is Not 1-preserving.
3. $A \bar{B} $ is Not Self-Dual. "$1$" is also Not Self-Dual.
4. $A \bar{B} $ is Not Linear.
5. $A \bar{B} $ is Not Monotonic.
So, by Post's theorem, $S$ is Functionally Complete.
Detailed Video Solution: https://www.youtube.com/watch?v=XmmkE4yRk5c&t=3795s
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