Can the negation of an implication statement be written in terms of implication operators? I know that $A\Rightarrow B$ is equivalent to $\neg A \lor B$. Also that $\neg (A\Rightarrow B)$ is given by $A \land \neg B$. But can $\neg (A\Rightarrow B)$ be written in terms of $A$, $B$, $\Rightarrow$, and $\neg$ alone (i.e. without the use of $\lor$ and $\land$)?  
 A: I think you are asking for the principal connective to be an implication rather than a negation. If that's right, a suitable equivalent to $\lnot (A \Rightarrow B)$ is $(A \Rightarrow B) \Rightarrow \lnot(A \Rightarrow A)$.
Note that it is impossible to write $\lnot(A \Rightarrow B)$ in terms of implication alone. If $\phi(A, B)$ is equivalent to $\lnot(A \Rightarrow B)$ , then $\phi(A \Rightarrow A, A)$ is equivalent to $\lnot A$, so as $\lnot$ and $\Rightarrow$ are functionally complete, if $\phi$ is constructed with $\Rightarrow$ alone, then $\Rightarrow$ on its own would be functionally complete, but by Post's characterisation of functionally complete sets of connectives, it is not (see link above).
A: I use Polish notation.
The formation rules go:


*

*All lower case letters of the Latin alphabet qualify as well-formed formulas.

*0 consists of a well-formed formula standing for the constant false proposition.

*If $\alpha$ and $\beta$ qualify as well-formed formulas, then so does C$\alpha$$\beta$.  C stands for implication, or equivalently conditional.


Now let us adopt the definition that the negation of a proposition Np, comes as abbreviation for Cp0, since C00 = 1 and N0 = 1, and C10 = 0 and N1 = 0.
Thus, NCab abbreviates CCab0.
More formally, we can take as a definition:


*

*C$\delta$Np$\delta$Cp0 where $\delta$ consists of a functioral variable of one argument (in other words $\delta$ comes as a variable which ranges over the four unary-truth functors of two-valued logic).


Substituting p with Cab and dropping $\delta$ or equivalently substituting $\delta$' with the apostrophe symbol ' where the apostrophe symbol ' consists of the argument of $\delta$ we have


*CNCabCCab0


as a thesis obtained from the definition.  Thus, given NCab by definition we obtain CCab0.  From 1. and the rule of uniform substitution for $\delta$ and the rule of uniform substitution for propositional variables we can also obtain NCab from CCab0.
