(p ⇒ q) ∧ (¬p ⇒ ¬q ∧r)
(q ∨r) ⇒ s
s ⇒ t
Therefore,
t
my efforts.
1 (p ⇒ q) ∧ (¬p ⇒ ¬q ∧r premise
2 (q ∨r) ⇒ s premise
3 s ⇒ t premise
4 p assumption
5 q from 1 & 4 ⇒ elim (modus ponens)
6 q∨r from 2 & 5, v Intro
7 s from 2 & 6, ⇒ elim (modus ponens)
8 t from 3 & 7, ⇒ elim (modus ponens)
9 ¬p assumption
10 ¬q ∧r from 1 & 9 ⇒ elim (modus ponens)
11 r from 10 ∧ Elim
12 q∨r from 11 ∨intro
13 s from 2 & 12, ⇒ elim (modus ponens)
14 t from 3 & 13, ⇒ elim (modus ponens)
1$\;(p \to q) \land (\lnot p \to (¬q \land r))$ $\quad$ premise
2$\;(q\lor r)\to s$ $\quad$ premise
3$\;s\to t$. $\quad $ premise
4 $\;(p\to q)\quad$ (1, $\land$-elim)
5 $\;\lnot p \to (\lnot q \land r)\quad (1, \land$-elim)
6 $\;p \lor \lnot p\quad$ (tautology: Assuming Law of the excluded middle)
7 $\qquad$ Assume $p.\;\;$ (assumption)
8 $\qquad\qquad q\quad $ from $4, 7$, elim modus ponens
9$\qquad\qquad q\lor r\quad $ from 8 $\lor$-intro
10 $\qquad\qquad s\quad$ from 2&9 elim modus ponens
11 $\qquad\qquad t\quad$ from 3&10 elim modus ponens
12 $\qquad p\to t\quad$ from $7-11$, $\to$-intro
10 $\qquad$ Assume $\lnot p\quad $ Assumption
11 $\qquad\qquad \lnot q\land r\quad$ 5, 10, modus ponens
12 $\qquad\qquad r\quad$ 11, $\land$-elim
13 $ \qquad \qquad q\lor r\quad $ 12 $\,\lor$-Intro
14$\qquad\qquad s\quad $ from 2, 13, modus ponens
15$\qquad\qquad t\quad$ from 3, 14, modus ponens
16 $\qquad\lnot p \to t\quad $ from 10-15, conditional introduction.
17 $\;(p\lor \lnot p)\to t$
18 $\;t$
Now, you can refer to your text and notes, and the rules of inference you've learned, to justify each step.