Translating "therefore" in propositional logic and using contradictions to prove an argument.

So I have the statements:

A Football player is either running at the oval or sitting in the library. If he is running, then he is not reading a book. The player is reading a book. Therefore, he is sitting in the library.

This is what I got so far:

Running at the Oval => R

Sitting in the Library=>S

Reading a Book=> B

Football player is running on the oval or sitting in the Library.

R V S

If Football player is running on the oval, then he is not reading a book

R→¬B

Football player is reading a book therefore he is sitting in the library.

From what I gather using "therefore" implies it's a material implication so:

B→S

If this is the case can I safely use the following to describe the statements given:

(R→¬B) V (B→S)

Solution 1:

"Football player is running on the oval or sitting in the Library.

R V S

If Football player is running on the oval, then he is not reading a book

R→¬B"

That works.

You didn't symbolize the sentence that the player is reading a book (ignoring the potential linguistic issue that a person reading a book is not a "player" of a sport at that time at least). That would be "B". So, the premises are:

(R V S)

R→¬B

Now, can you deduce the conclusion S?

...

Spoiler:

B is true. So, $\lnot$B is false. (R→¬B) is true, so that's (R$\rightarrow$F). (R$\rightarrow$F) is only true when R is F. (R$\lor$S) is true, so we have (F$\lor$S). But, that is only true when S is true. So, the person is sitting in the library. Thus, the argument is valid. Supposing it were invalid would be a contradiction. So, the argument is valid.

Solution 2:

The word ‘therefore’ $(\therefore)$ translates neither as ‘implies’ $(\implies)\;$ nor as the material conditional/implication $(\rightarrow).$

The meta-sentence “$P$ is true; therefore $Q$ is true” $(P;\;\therefore Q)$ has a different meaning from the meta-sentence “if $P$ is true, then $Q$ is true” $(P\implies Q):$ the former implicitly contains the latter (which doesn't indicate whether $P$ is true), and additionally asserts that $P$ is true.

(The meta-sentence $\,P\implies Q\,$ just means that the sentence $\,P\rightarrow Q\;$ (“if $P,$ then $Q$”) is true. The rest of this answer ignores this nuance.)
In an argument, the word ‘therefore’ indicates that the statements before it are (true) premises, and asserts that the statement after it is the conclusion. So, your given argument contains $3$ premises $$R \lor S\\ R \rightarrow\lnot B\\ B$$ and a conclusion $$S.$$
An argument is valid if and only if its conclusion is inevitable, given its premises. If we first assume (contrarily^‡) that its conclusion is false, and then, from its premises, successfully derive some contradiction (absurdity or logical falsity), we will have proven that the argument is valid.

So, for our example, start by assuming that $S$ is false. Then, by Premise 1, $R$ must be true. Then, by Premise 2, $B$ must be false. But by Premise 3, $B$ is true. We have derived the absurdity $$B \land \lnot B,$$ which must have arisen from our assumption. Thus, $S$ must instead be true. Hence, the argument is indeed valid.

$\quad$^{^‡ The word "contradiction" does not refer to this step's negating of the conclusion, nor does a proof by contradiction generally involve contradicting a premise (in which case the proof can be constructed more succinctly as a proof by contrapositive).}