I've never actually solved a problem like this before, but it looks pretty trivial so I'll give it a shot. My apologies if this is wrong.
i. Let $\Phi$ denote the formulae of propositional logic that can be formed from the connectives in $C$ and the variables in $P$. More precisely, lets us defined that $\Phi$ is the smallest collection of formulae such that
- If $X \in P$, then $X \in \Phi$
- If $\phi,\psi \in \Phi$ then $\phi \wedge \psi \in \Phi$.
- If $\phi,\psi \in \Phi$ then $\phi \vee \psi \in \Phi$.
Furthermore, let $I$ denote a valuation such that $I \models X$ for every $X \in P$, and let $\Phi'$ denote the set of all formulae $\phi$ of propositional logic such that $I \models \phi$. The problem becomes:
Show that $\Phi \subseteq \Phi'$.
Now for the important realization:
Since $\Phi$ is the least set satisfying 1,2 and 3, thus it suffices to show that $\Phi'$ also satisfies 1,2 and 3.
That's it, the rest is easy. We continue:
In other words, it suffices to show the following.
- If $X \in P$, then $I \models X$.
- If $I \models \phi,\psi$, then $I \models \phi \wedge \psi$.
- If $I \models \phi,\psi$, then $I \models \phi \vee \psi$.
But this is trivial.
- True, because we assumed that $I \models X$ for every $X \in P.$
- True, by the definition of $\wedge$.
- True, by the definition of $\vee$.
ii. I'm not sure what the definition of "adequate" is, but I'm guessing this is even easier. If it just means: "can be used to express all functions of the form $\mathbb{B}^n \rightarrow \mathbb{B},$" well just take any function returning "FALSE" whenever all arguments are true and you'll have your counterexample.