Natural deduction proof of $p \rightarrow q \vdash \lnot(p \land \lnot q)$ So yeah, the entire question is pretty much in the title. 
$$p \rightarrow q \vdash \lnot(p \land \lnot q)$$
I've been able to derive the reverse, but I don't how to logically go from the premise to the conclusion using natural deduction only. I can see that the two formulas are equal using transformations.
These are the rules I'm allowed to use:

Please help me understand how to do this. 

$1.$ $p \rightarrow q  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  $       -- (Premise)  
$2.$ $p \wedge \neg q \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $   -- (Assume the contrary to what has to be proved in the conclusion)  
$3.$ $p \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ -- ($\wedge E$ on $2.$)   
$4.$ $\neg q \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ -- ($\wedge E$ on $2.$)   
$5.$ $q \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  $ -- (Modus ponens on $1.$ and $3.$)   
$6.$ $\bot \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ -- (bot introduction due to contradiction on $4.$ and $5.$)   
$7.$ $\neg(p \wedge \neg q) \ \ \ \ \ \ \ \ \ \ \  $   --(assumption wrong due to arrival of contradiction) 
