Expressive adequacy After reading a few others posts on this site, I am still struggling in showing how sets are incomplete.
(a) Show that {$↔,\bot, ∧$} is functionally complete, but that no proper subset is.
(b) is solved! :)
(b) Assume that $c$ is a 2-place connective. Show that if either $f_c(\top, \top) = \top$ or $f_c(\bot, \bot) = \bot$, then {$c$} is not complete. 
For (b), I understand that for {$\uparrow$} and {$\downarrow$}, the only functionally complete 2-place connectives, that $f_{\uparrow}(\top, \top) = \bot$ and $f_{\uparrow}(\bot, \bot) = \top$. The same results apply for $f_{\downarrow}$. I just can't see how to then apply this to prove $f_c(\top, \top) = \top$ means {$c$} is incomplete.

For (a), to prove that a set of connectives is functionally complete, an "induction on formulas" should be done by demonstrating that the other known connectives in the logic can be derived using our given set. In our case, the other connectives to be derived are $\{\vee,\top,\rightarrow,\neg\}$ or just $\{\vee, \top, \neg\}$ because $\rightarrow$ can be derived already using $\vee$ and $\neg$ $(p\rightarrow q=\neg p\vee q)$.
The formula $\neg p$ can be derived using $p\leftrightarrow\bot$ because when $p$ is $\top$, "$p\leftrightarrow\bot$" gives $\top\leftrightarrow\bot=\bot$ and when $p$ is $\bot$, "$p\leftrightarrow\bot$" gives $\bot\leftrightarrow\bot=\top$.
It's now obvious that $\top$ can be derived using $\bot\rightarrow\bot$.
Finally, $p\vee q$ can also be derived from $\{\leftrightarrow,\bot,\wedge\}$. Hint: De Morgan's Theorem.
