Help Proving this Propositional logic Can someone help me with my proof?
Let p, q, r and s be propositions. Consider the hypothesis $(p\space\lor\space q \to r)$, $\lnot s$, $r \to s$ and conclude $\lnot p$.
My Proof
$
r \to s\\
\lnot r \to s\\
\lnot r\space\lor\space p\space\lor\space q\\  
$ 
This is all I have so far
 A: Hint 1: $(r \to s) \leftrightarrow (\lnot s \to \lnot r)$
Hint 2: $\lnot(p\space\lor\space q) \leftrightarrow (\lnot p\space\land\space \lnot q) $
A: $$(((p\lor q) \to r)\land\lnot s\land (r \to s))\to\lnot p$$
$$\equiv(((p\lor q) \to r)\land\lnot s\land (\lnot s \to \lnot r))\to\lnot p$$
$$\equiv(((p\lor q) \to r)\land\lnot r)\to\lnot p$$
$$\equiv(\lnot r\land(\lnot r\to\lnot(p\lor q)))\to\lnot p$$
$$\equiv\lnot(p\lor q)\to\lnot p$$
$$\equiv p\to (p\lor q)$$
$$\equiv \lnot p\lor p\lor q$$
$$\equiv \top$$
A: Unfortunately, your "proof" isn't even a partial proof. You don't explain how you reach each step after the first, and you seem to be applying "rules" that don't exist (or misapplying rules you misunderstood)
The 2nd step is not valid inference: what "inference rule" do you think you're using there? Together with $r\to s$, we can conclude that $s$. But we also have $\neg s$; hence, a contradiction (from which everything follows, including $\neg p$... but that's cheating). 
How did you get the 3rd step? It looks like you're misapplying the rule $A\to B\equiv \neg A\lor B$, using what would the rule for reverse implication $A\leftarrow B \equiv B\to A \equiv A\lor \neg B$. 
Because $A\to B \equiv \neg B \to \neg A$, a valid rule of inference (modus tollens) is:
$$
\text{from $A\to B$ and $\neg B$, conclude $\neg A$}.
$$
You can use this twice in a row, starting from your hypotheses. You'll also need to know that $\neg\:(A\lor B) \equiv (\neg A \land \neg B)$
A: That is not a good start.   There's no justification for any of your steps, and they cannot be obtained from the premises by any valid rules of inference.   Try again.
Here's an outline of a proof.   Can you supply the missing details?
$$\begin{align}\because~&(p\vee q)\to r, \neg s, r\to s \\ \hdashline&{\begin{array}{l|l:l}1 & (p\vee q)\to r & \textsf{Premise} \\ 2&\neg s& \textsf{Premise}\\3&r\to s& \textsf{Premise} \\[3ex] & \neg r & \textit{why?} \\[3ex]  & \neg(p\vee q) & \textit{why?} \\[3ex] & \neg p \end{array}} \\ \hline \therefore~ &~ \neg p  \end{align}$$
Consider what rules of inference you have available.
