This is a simple application of Post’s characterization of functional completeness.
Here is an example in three variables below.
Let ${\cal F}=\lbrace f_1,f_2,f_3,f_4 \rbrace$ with
\begin{align}
f_1&=0,\\
f_2&=1,\\
f_3(x_1,x_2,x_3)&=x_1\oplus x_2 \oplus x_3,\\
f_4(x_1,x_2,x_3)&=x_1 \wedge x_2 \wedge x_3
\end{align}
Then :
$\lbrace f_1,f_2,f_3 \rbrace$ generates only affine boolean functions,
so this set is not functionally complete.
$\lbrace f_1,f_2,f_4 \rbrace$ generates only monotone boolean functions,
so this set is not functionally complete.
$\lbrace f_1,f_3,f_4 \rbrace$ generates only falsity-preserving boolean functions,
so this set is not functionally complete.
$\lbrace f_2,f_3,f_4 \rbrace$ generates only truth-preserving boolean functions,
so this set is not functionally complete.
On the other hand, $\cal F$ is functionally complete (indeed, $\neg x_1$ can be
expressed as $f_3(x_1,f_1,f_2)$ and $x_1 \wedge x_2$ as $f_4(f_2,x_1,x_2)$), so it is an irreducible functionally complete set with four functions.
Note we have $x \land \neg y = x - x \land y$. Using this fact
$$(x \land y) (x \lor y) + (x \land \neg y) (\neg x \land y) = x y$$
becomes
$$(x \land y) (x \lor y) + (x - x \land y) (y - x \land y) = x y$$
expanding,
$$(x \land y) (x \lor y) + x y - (x + y) (x \land y) + (x \land y)^2 = x y$$
$$(x \land y) (x \lor y) - (x + y) (x \land y) + (x \land y)^2 = 0$$
Divide by $x \land y$:
$$(x \lor y) - (x + y) + (x \land y) = 0$$
$$(x \lor y) + (x \land y) = x + y$$
To prove this equality, write down $x$ and $y$ in binary: $x = \sum_{i=0}^{n-1} x_i 2^i$ and $y = \sum_{i=0}^{n-1} y_i 2^i$, where $x_i, y_i \in \{0,1\}$. Note that $x \land y = \sum_{i=0}^{n-1} \min (x_i, y_i) 2^i$ and $x \lor y = \sum_{i=0}^{n-1} \max (x_i, y_i) 2^i$. We want to show
$$\sum_{i=0}^{n-1} 2^i \max (x_i, y_i) + \sum_{i=0}^{n-1} 2^i \min (x_i, y_i) = \sum_{i=0}^{n-1} 2^i x_i + \sum_{i=0}^{n-1} 2^i y_i$$
which is the same as
$$\sum_{i=0}^{n-1} 2^i (\max (x_i, y_i) + \min (x_i, y_i)) = \sum_{i=0}^{n-1} 2^i (x_i + y_i)$$
but for any numbers $x_i, y_i$ we have $\max (x_i, y_i) + \min(x_i, y_i) = x_i + y_i$.
Best Answer
Here are two approaches to proving that negation and majority don't form a complete set.
(1) Both commute with negation. In particular, $\text{Maj}(\neg x,\neg y,\neg z)=\neg \text{Maj}(x,y,z)$. Deduce that the same property holds for any Boolean operation obtainable by composing these.
(2) You can write out explicitly a complete list of the functions of two variables $x,y$ obtainable by iterating $\neg$ and Maj. They are simply $x,\neg x, y, \neg y$. Check that anything obtained by applying $\neg$ to any of these or applying Maj to any of these is again one of these. In particular, you cannot get $x\land y$ this way.