How to prove that set $\{\wedge, \vee \}$ is not functional complete?
I tried to do it like this, I tried to show that $\neg x$ couldnt be received by $\wedge$ and $\vee$.
I take function $\neg x = \phi(x)$. Here I want to show that $\neg x \neq \phi(x)$. And after that I use induction.
Step $k=0$: $x = \phi(x)$, if $x=0$ then $\phi(0) = 0$. After that $\neg x \neq x$, because $0 \neq 1$.
Step $k = 1$: $\phi(x) = x \wedge x$ or $\phi(x) = x \vee x$ and here I have problems. I dont know what to do next...
In stead of focusing on a function that takes just one variable, it may be a little easier to think of a function with two variables, just like the $\land$ and $\lor$.
In particular, show (using induction!) that using $\land $ and $\lor$ alone, you cannot capture any function of two variables that returns False when both variables are True, for example $\neg (p \land q)$.