Prove that $\lnot (p \implies q)$ is equivalent to $p \land \lnot q$? By equivalent I mean the biconditional, as in
$$\lnot (p \implies  q) \iff p \land \lnot q$$
Given the definition of implication, I understand why this is true, but I need a bit of help showing this with a formal proof using rules like $\lor-\text{Elim}$  and $\bot-\text{Intro}$. 

$$\newcommand{\DeductionBox} [1]{\begin{array} {l|} #1 \\ \hline \end{array}}$$
The reverse direction is straightforward:
$$\begin{array} {l}
\begin{array} {rlr}
(1) & p \land \lnot q             & \text{Given} \\
(2) & p                           & \land\text{-Elimination of }(1) \\
(3) & \lnot q                     & \land\text{-Elimination of }(1) \\
\end{array}\\
\\
\DeductionBox{
  \begin{array} {rlr}
  \quad (4) & p \implies  q      &  \text{New Assumption}              \\
  \quad (5) & q                  &  \implies\text{-Elimination of }(4) \text{ and } (2) \\
  \quad (6) & \bot               &  \bot\text{-Introduction of }(5) \text{ and }(3)
  \end{array} \\
}  \\
\\
\begin{array} {rlr}
(7) & (p \implies q) \implies \bot & \quad\implies-\text{Introduction of }(4) \text{ to } (6)   \\
(8) & \lnot (p \implies q) & \lnot-\text{Introduction of }(7)                               \\
\end{array}
\end{array}$$
$$~$$
$$~$$
The foward direction can be done using the law of the excluded middle:
$$\begin{array} {l}
\begin{array} {rlr}
(1) & \lnot (p \implies q) & \text{Given} \\
(2) & \bot \implies \lnot q & \text{Vaccous Implication } \\
\end{array}\\
\\
\DeductionBox{
  \begin{array} {rlr}
  \quad (3) & p      &  \text{New Assumption}   \\
  \end{array} \\
  \\
  \DeductionBox{
    \begin{array} {rlr}
    \quad \quad (4) & q      &  \text{New Assumption}   \\
    \end{array} \\
    \\    
    \DeductionBox{
      \begin{array} {rlr}
      \quad \quad \quad (5) & p      &  \text{New Assumption}   \\
      \quad \quad \quad (6) & q      &  \text{Copy of }(4)   \\
      \end{array} \\
    }\\
    \\
    \begin{array} {rlr}
    \quad \quad (7)  & p \implies q          & \implies\text{-Introduction of }(5) \text{ to }(6) \\
    \quad \quad (8)  & \bot                  & \bot\text{-Introduction of }(7) \text{ and }(1) \\
    \quad \quad (9)  & \lnot q               & \implies\text{-Elimination of }(2) \text{ and }(8) \\
    \quad \quad (10) & p \land \lnot q       & \land\text{-Introduction of }(3) \text{ and }(9) \\
    \end{array} \\
  }\\
  \\  
  \DeductionBox{
    \begin{array} {rlr}
    \quad \quad (11) & \lnot q      &  \text{New Assumption}   \\
    \quad \quad (12) & p \land \lnot q & \land\text{-Introduction of }(3)\text{ and }(11) \\
    \end{array} \\
  }\\
  \\  
  \begin{array} {rlr}
  \quad (13) & q \implies (p \land \lnot q)        &  \implies\text{-Elimination of }(4)\text{ to }(10) \\
  \quad (14) & \lnot q \implies (p \land \lnot q)  &  \implies\text{-Elimination of }(11)\text{ to }(12) \\
  \quad (15) & q \lor \lnot q                      &  \text{Law of the Excluded Middle}   \\
  \quad (16) & p \land \lnot q                     &  \lor\text{-Elimination of }(15),(13),(14)   \\
  \end{array} \\
} \\
\\
\DeductionBox{
  \begin{array} {rlr}
  \quad (17) & \lnot p         &  \text{New Assumption } \\
  \end{array}\\
  \\
  \DeductionBox{
    \begin{array} {rlr}
    \quad \quad (18) & p                 & \text{New Assumption} \\
    \quad \quad (19) & \bot              & \bot\text{-Introduction of }(18)\text{ and }(17) \\
    \quad \quad (20) & \bot \implies q   & \text{Vaccuous Implication}  \\
    \quad \quad (21) & q                 & \implies\text{-Elimination of }(20)\text{ and }(19)   \\
    \end{array} \\
  }\\
  \\
  \begin{array} {rlr}
  \quad (22) & p \implies q     & \implies\text{-Introduction of }(18)\text{ to }(21) \\
  \quad (23) & \bot             & \implies\text{-Elimination of }(1)\text{ and }(22) \\
  \quad (24) & \lnot q          & \implies\text{-Elimination of }(2)\text{ and }(23) \\
  \quad (25) & p \land \lnot q  & \land\text{-Introduction of }(3)\text{ and }(24) \\
  \end{array}\\
}\\
\begin{array} {rlr}
(26) & p \implies (p \land \lnot q)        &  \implies\text{-Elimination of }(3)\text{ to }(16) \\
(27) & \lnot p \implies (p \land \lnot q)  &  \implies\text{-Elimination of }(17)\text{ to }(25) \\
(28) & p \lor \lnot p                      &  \text{Law of the Excluded Middle}   \\
(29) & p \land \lnot q                     &  \lor\text{-Elimination of }(28),(26),(27)   \\
\end{array}\\
\end{array}$$
