First we prove that
$$
\{\neg(r\to p)\lor[(\neg q\to\neg p)\land(r\to q)]\}\iff \neg p \lor q
$$
Let $A\iff\{\neg(r\to p)\lor[(\neg q\to\neg p)\land(r\to q)]\}$. Then
\begin{align}
A&\iff\{\neg(\neg r\lor p)\lor[(\neg q\to\neg p)\land(r\to q)]\}\tag1
\\
&\iff\{(r\land\neg p)\lor[(\neg q\to\neg p)\land(r\to q)]\}\tag2
\\
&\iff\{(r\land\neg p)\lor[(p\to q)\land(r\to q)]\}\tag3
\\
&\iff\{(r\land\neg p)\lor(p\to q)\}\land\{(r\land\neg p)\lor(r\to q)\}\tag4
\\
&\iff\{(r\land\neg p)\lor(\neg p\lor q)\}\land\{(r\land\neg p)\lor(\neg r\lor q)\}\tag5
\\
&\iff\{[r\lor(\neg p\lor q)]\land[\neg p\lor(\neg p\lor q)]\}\land\{[r\lor(\neg r\lor q)]\land[\neg p\lor(\neg r\lor q)]\}\tag6
\\
&\iff\{[r\lor(\neg p\lor q)]\land[\neg p\lor q]\}\land\{[(r\lor(\neg r)\lor q]\land[\neg p\lor(\neg r\lor q)]\}\tag7
\\
&\iff\{[r\lor(\neg p\lor q)]\land[\neg p\lor q]\}\land\{1\land[\neg p\lor(\neg r\lor q)]\}\tag8
\\
&\iff\{[r\lor(\neg p\lor q)]\land[\neg p\lor(\neg r\lor q)]\}\land[\neg p\lor q]\tag9
\\
&\iff\{[r\lor(\neg p\lor q)]\land\{\neg r\lor(\neg p\lor q)\}\land[\neg p\lor q]\tag{10}
\\
&\iff\{0\lor(\neg p\lor q)\}\land[\neg p\lor q]\tag{11}
\\
&\iff(\neg p\lor q)\land(\neg p\lor q)\tag{12}
\\
&\iff\neg p\lor q
\end{align}
$(1)$ is for $r\to p\iff \neg r\lor p$.
$(2)$ is for De Morgan's law and double negation.
$(3)$ is for $(\neg q\to\neg p)\iff (p\to q)$.
$(4)$ is for distributivity.
$(5)$ is for $r\to q\iff \neg r\lor q$.
$(6)$ is for distributivity.
$(7)$ is for idempotence of disjunction and associativity.
$(8)$ is for $r\lor \neg r\iff 1$.
$(9)$ is for $1\land B\iff B$.
$(10)$ is for associativity and commutativity.
$(11)$ is for distributivity and $r\land \neg r\iff 0$.
$(12)$ is for idempotence of conjunction.
So
$$
\{\{\neg(r\to p)\lor[(\neg q\to\neg p)\land(r\to q)]\}\to(\neg p\lor q)\}\iff\{(\neg p\lor q)\to(\neg p\lor q)\}
$$
Since
\begin{align}
\{(\neg p\lor q)\to(\neg p\lor q)\}&\iff\{\neg[(\neg p\lor q)]\lor(\neg p\lor q)\}
\\
&\iff\{(p\land \neg q)\lor(\neg p\lor q)\}
\\
&\iff\{[(p\lor(\neg p\lor q)]\land [\neg q\lor(\neg p\lor q)]\}
\\
&\iff\{[(p\lor\neg p)\lor q)]\land [(\neg q\lor q)\lor \neg p)]\}
\\
&\iff\{(1\lor q)\land (1\lor \neg p)\}
\\
&\iff\{1\land 1\}
\\
&\iff 1
\end{align}
$\{(\neg p\lor q)\to(\neg p\lor q)\}$ is tautology. So
$$
\{\neg(r\to p)\lor[(\neg q\to\neg p)\land(r\to q)]\}\to(\neg p\lor q)\iff 1
$$
It is tautology too.