Prove that the conective $\beta(a,b,c,d)=(a\implies b) \implies (c\implies d)$ is not adequate. A set C of conectives is called adequate if you every formula written with the conectives $\{\text{implies, and, not, or}\}$ has an equivalent formula written only with conectives from $C$.
Prove that the conective $\beta(a,b,c,d)=(a\implies b) \implies (c\implies d)$ is / is not not adequate.
I believe this conective is not adequate: I've come to think you can't form contradictions using it alone. Also, I could form the $\implies$ conective using $\beta$, thus $\{\beta, \implies\} \equiv\{\beta\}$ in terms of "formula making power".
E: I should add I haven't yet studied boolean algebras, so I really don't understand many terms.
Is it possible to make a (perhaps sillier) argument by contradiction, assuming there exists a contradiction expressible only with $\beta$ and reaching and absurdity?
 A: Yes, implication can be expressed by $\beta$, as $\ a\Rightarrow a\,=\,\top$ (true) always, and thus
$$a\Rightarrow b\ =\ \beta(a,a,a,b)$$
Also, as $\beta$ is composed only of implications, we have that the expressive power of $\beta$ is just the same as that of $\Rightarrow$.
Now, to see that the constant falsum $\bot$ can't be expressed, we can consider a map $f$ between Boolean algebras $\{\bot,\top\}\to\{\bot,\,x,\,\lnot x,\,\top\}$ sending $\bot\mapsto x$ and $\top\mapsto\top$. 
This map preserves $\Rightarrow$, thus also $\beta$, but doesn't preserve the zero element $\bot$.
A: Berci’s proof is elegant; here’s a less elegant but perhaps slightly more elementary one. Start with the same observation, that the expressive power of $\beta$ is the same as that of $\Rightarrow$. Now suppose that we can express $\bot$ using only $\Rightarrow$; then among all expressions that do this, there must be one of shortest possible length. Since $\Rightarrow$ is the only available connective, this shortest formula must have the form $\varphi\Rightarrow\psi$ for some shorter formulas $\varphi$ and $\psi$. This formula $\varphi\Rightarrow\psi$ is false for all truth assignments to the proposition letters appearing in it, which means that $\varphi$ is always true, and $psi$ is always false. But that means that $\psi=\bot$, where $\psi$ is shorter than $\varphi\Rightarrow\psi$, contradicting the minimality of $\varphi\Rightarrow\psi$. Thus, there is no formula using only $\Rightarrow$ that expresses $\bot$, and $\{\Rightarrow\}$ (and hence $\{\beta\}$) is not complete.
A: Post's characterisation of functional completeness gives necessary and sufficient conditions for the adequacy of a set of propositional connectives. $\{\beta\}$ fails Post's test because $\beta(\top, \top, \top, \top) = \top$.
