Answer
First we prove that premises $(p \land t) \rightarrow (r \lor s), q \rightarrow (u \land t), u \rightarrow p, \lnot s$ and $q$ leads to the conclusion $r $. This will prove that same premises (excluding q ) leads to the conclusion $q\rightarrow r$ through exercise 11. Then we prove the same with rules of inference.
Work Step by Step
A.) Through exercise 11
1.) $q$
2.) $q \rightarrow (u \land t)$
3.) $(u \land t)$ {Modus ponens from 1.) and 2.)}
4.) $u$ {Simplification from 3.)}
5.) $u \rightarrow p$
6.) $p$ {Modus ponens for 4.) and 5.)}
7.) $t$ {Simplification from 3.)}
8.) $p \land t$ {Conjunction from 6.) and 7.)}
9.) $(p \land t) \rightarrow (r \lor s)$
10.) $r \lor s$ {Modus ponens from 8.) and 9.)}
11.) $\lnot s$
12.) $r$ {Disjunctive Syllogism from 10.) and 11.)}
We have proved that premises $(p \land t) \rightarrow (r \lor s), q \rightarrow (u \land t), u \rightarrow p, \lnot s$ and $q$ leads to conclusion $r$. Hence by proof in exercise 11 we can say that the same premises excluding $q$ leads to conclusion $q
\rightarrow r$.
B.) Through rules of inference
1.) $q \rightarrow (u \land t)$
2.) $q \rightarrow u$ {From 1.) equivalence of conditional statements}
3.) $q \rightarrow t$ {From 1.) equivalence of conditional statements}
4.) $u\rightarrow p$
5.) $q\rightarrow p$ {Hypothetical Syllogism from 2.) and 4.)}
6.) $q\rightarrow (p \land t)$ {From 3.) and 5.) equivalence of conditional statements}
7.)$(p \land t) \rightarrow (r \lor s)$
8.) $q\rightarrow (r \lor s)$ {Hypothetical syllogism from 6.) and 7.) }
9.) $\lnot s$
10.) $s\equiv F$ {Laws of negation from 9.)}
11.) $p \rightarrow r $ {Identity Law from 10.) and 8.) }
Hence, we have proved with the rules of inference that premises $(p \land t) \rightarrow (r \lor s), q \rightarrow (u \land t), u \rightarrow p, \lnot s$ leads to conclusion $p \rightarrow r $
