Proving conclusion c when give premises I am given three premises:
P1: $ r \implies s $
P2: $ p \implies (q \land r) $
P3: $ \neg (q \land r) $
and need to get to the conclusion:
C: $ \neg p $
I am having trouble with this question because I don't know where to begin with the given premises. If there were a premise with an $ \land $ in it (i.e. $ p \land r $) because then I can either assume p or r to be true. But in this case, I am not able to assume any one variable, or rather, don't know how to. Can someone help me out?
 A: Also, P2 p⟹(q∧r) is equivalent to ¬(q∧r)⟹¬p.  Since you have P3 ¬(q∧r), you have immediately ¬p.  You do not need P1 r⟹s.
A: The statement $P\Rightarrow Q$ is logically equivalent to $Q\vee\neg P$ (this can be confirmed with truth tables). Applying this to premise two, we get $(q\wedge r)\vee\neg p$. By premise three, we know $q\wedge r$ is false, so $\neg p$ must be true. I'm not sure why premise one is there, it is not necessary for the proof.
A: Hint: Part of the exercise is to identify which premises are useful.   You donut have to use them all.

But in this case, I am not able to assume any one variable, or rather, don't know how to. 

Just do it.
Consider a proof by contradiction.   Assume $p$ and demonstrate that contradicts some of the premises.   Therefore those premises infer $\neg p$.
$$\begin{array}{l|l} 1 & r\to s  &\text{premise }1\\ 2 & p\to (q\wedge r)&\text{premise }2\\ 3 & \neg(q\wedge r)&\text{premise }3
\\ \quad 4.1 & p &\text{assume }p\\ \quad \vdots & \vdots &\vdots \\ \quad 4.x & \bot & \text{reasons} \\ 5 & \neg p & 4.1,4.x\text{ contradiction}  \end{array}$$
