A complement to the comments above.
It's worth mentioning that the meaning of the propositional connectives $\neg$, $\wedge$, $\vee$, $\to$ should not be regarded as a mere symbolic translation of the meaning of their English counterparts "not", "and", "or", "if ... then" respectively. See Hedman's A First Course in Logic (2004), p.1-2:
Unlike their English counterparts, these symbols represent concepts that are precise and invariable. The meaning of an English word, on the other hand, always depends on the context. For example, ∧ represents a concept that is similar but not identical to “and.” For atomic formulas A and B, A ∧ B always means the same as B∧A. This is not always true of the word “and.” The sentence
She became violently sick and she went to the doctor.
does not have the same meaning as
She went to the doctor and she became violently sick.
Likewise ∨ differs from “or.” Conversationally, the use of “A or B” often precludes
the possibility of both A and B. In propositional logic A∨B always means
either A or B or both A and B.
This is the case in the sentences above:
(1) Catch Billy a fish, and you will feed him for a day.
(2) Teach him to fish, and you'll feed him for life.
Note that the "and" here should not be interpreted as "$\wedge$". We have many similar cases:
(i) Jump and you die
We intuitively know that this sentence actually means:
(i') If you jump then you will die
Hence the argument is stated this way:
- $C \to D$
- $T \to L$
$\therefore \neg L \vee T$
Which is not valid (note that the conclusion says '$L \to T$').
\begin{align} \lnot p \vee q \rightarrow r \tag{1}\\
s \vee \lnot q \tag{2} \\
\lnot t \tag{3}\\
p \rightarrow t \tag{4}\\
\lnot p \wedge r \rightarrow \lnot s \tag{5}\end{align}
As you note in your post, from the fourth premise we have $p\to t$, and we have, in the third premise, $\lnot t$. By modus tollens, we derive $\lnot p$.
Now, since we've derived $\lnot p$ from premises (3), (4), we also have $\lnot p \lor q$, by "addition" to $\lnot p$, (also called "or-introduction" which is shorthand for disjunction introduction).
And from $\lnot p \lor q$, together with the premise (1), we have $r$ by modus ponens.
Now, since we already deduced $\lnot p$, and we just deduced $r$, we can use "And-introduction (conjunction-introduction)" to get $\lnot p \land r$.
Given $\lnot p \land r$, and premise (5): $(\lnot p \land r) \to \lnot s$, we have, by modus ponens, $\lnot s$.
But given our premise (2), $s \lor \lnot q,$ together with $\lnot s$, we deduce $\lnot q$, as desired.
Best Answer
The argument is valid. You can see it informally as follows. If $t$ were false, then both $s$ and $t$ would be false. Since $r\to(s\lor t)$, this means that $r$ must be false, and since $p\to(r\land q)$, this in turn means that $p$ must be false. But then $p\land q$ is false, contrary to the first assumption.
If all else fails, you can work it out by a truth table, verifying that in every line in which all four of the hypotheses are true, $t$ is true. We don’t even have to look at all $32$ possible lines: from the hypothesis $p\land q$ we know that $p$ and $q$ are true, and from the hypothesis $\neg s$ we know that $s$ is false. Thus, only $r$ and $t$ are free to vary, and we have only four lines to consider:
$$\begin{array}{ccc} p&q&s&r&t&r\land q&p\to(r\land q)&s\lor t&r\to(s\lor t)\\ \hline T&T&F&T&\color{brown}T&T&\color{brown}T&T&\color{brown}T\\ T&T&F&T&F&T&T&F&F\\ T&T&F&F&T&F&F&T&T\\ T&T&F&F&F&F&F&F&T \end{array}$$
The only row in which the second and third hypotheses are both true is the first, and in it $t$ is also true.