Given the Premises $P,\lnot Q \implies \lnot P,$ and $ Q\lor R$ prove $\lnot R$
Ok so my proof looks like this so far: $$P...Premise$$ $$P \implies Q ... Contrapositive$$ $$Q\lor R...Premise$$
But I am stuck here I do not know how to reach from $R$ to $\lnot R$. Also I am unsure the notation but the $\lor$ should be and "exclusive or".
Let's use $\oplus$ to mean "exclusive or" instead of the logical "$\lor$".
We define $$A \oplus B = (A\lor B) \land \lnot (A\land B) \equiv (A\land \lnot B) \lor (\lnot A \land B)$$
So we have the following premises:
$P\tag 1$
$P\rightarrow Q\tag 2$
$Q\oplus R\tag 3$
From premises (1) and (2), we can use modus ponens to conclude $$Q \tag 4$$
Now, from the definition of the exclusive-or above, $$Q \land (Q \oplus R) \tag{(3), (4)}$$
imply $\lnot R\tag{5}$
because $Q\oplus R = (Q\land \lnot R) \lor (\lnot Q \land R)\tag 3$
together with $Q = \lnot\lnot Q$ means that $(\lnot Q \land R)$ is false, giving us the use of the disjunctive syllogism to infer, therefore $Q \land \lnot R$, and using $\land$-elimination, we have $\lnot R$, as desired.
The or has to be exclusive for your statement to be true. An exclusive or is of the form $(Q \vee R) \wedge \neg(Q \wedge R)$, where $\vee$ is inclusive or. Now we have the following proof tree: $$\cfrac{\cfrac{\cfrac{\cfrac{P \quad \cfrac{\neg Q \rightarrow \neg P}{P \rightarrow Q}}{Q}\rightarrow E \qquad {}^\dagger R^1}{Q \wedge R}\wedge I\quad \cfrac{(Q \vee R) \wedge \neg(Q \wedge R)}{\neg(Q \wedge R)}\wedge E}{\bot}\neg E}{\neg R}\neg I, 1$$
$\rightarrow E$ is implication elimination, $\wedge I$ is and introduction, $\wedge E$ is and elimination, $\neg E$ is negation elimination and $\neg I$ is negation introduction. The ${}^\dagger$ denotes a marked assumption. More information can be found in Sets, Models and Proofs.