I claim that we can get rid of either ∨ or ∧, and the remaining set of {¬,∨,∧} is still functionally complete. This can be proven by de Morgan’s law.
A∨B = ¬(¬A∧¬B), therefore ∨ is not necessary when ¬ and ∧ are available. Likewise, A∧B = ¬(¬A∨¬B), therefore ∧ is not necessary when ¬ and ∨ are available.
Consequently, {¬,∧} and {¬,∨} are both minimum functionally equivalent sets.
In computer and electrical engineering, or-gates and and-gates are not that “natural”. Instead, nand-gates and nor-gates are more natural (due to the way semiconductor devices operate). Let ⋅ represent nand, and + represent nand and nor, respectively. Then A⋅B = ¬(A ∧ B), and A+B = ¬(A ∨ B). I claim that {⋅} and {+} are both minimum functionally complete sets.
Here is the proof. ¬A = A+A = A⋅A, and A∧B = (A⋅B)⋅(A⋅B) = (A+A)+(B+B). Because we already know that {¬,∧} and {¬,∨} are both minimum functionally complete sets, it follows that {⋅} and {+} are functionally complete, as well.