Did I Write this Statement Form in Polish Notation Correctly? My goal is the write the statement form in the first line in Polish notation. The last line is my solution. The intermediate steps show my work.

\begin{aligned}
     &((((A\Rightarrow B)\Rightarrow(C\Rightarrow D))\land(\lnot A))\lor C)\\      &\lor(((A\Rightarrow B)\Rightarrow(C\Rightarrow D))\land(\lnot A))C\\
     &\lor(((A\Rightarrow B)\Rightarrow(C\Rightarrow D))\land\lnot A)C\\
     &\lor(((\Rightarrow AB)\Rightarrow(\Rightarrow CD))\land\lnot A)C\\
     &\lor((\Rightarrow(\Rightarrow AB)(\Rightarrow CD))\land\lnot A)C\\
     &\lor((\Rightarrow\Rightarrow AB\Rightarrow CD)\land\lnot A)C\\
     &\lor(\land\Rightarrow\Rightarrow AB\Rightarrow CD\lnot A)C\\
     &\lor\land\Rightarrow\Rightarrow AB\Rightarrow CD\lnot AC\\
     \end{aligned}
(This is part of exercise 1.18(c) in Elliott Mendelson's Introduction to Mathematical Logic, Sixth Edition. I am self-studying that book.)
 A: The conversion into Polish notation is correct.
Let us record the conventional letters of Polish notation for reference:
\begin{alignat*}{3}
&Negation  &\qquad  \neg\varphi &\qquad  \mathrm{N}\varphi\psi\\
&Conjunction  &\qquad \varphi\wedge\psi &\qquad \mathrm{K}\varphi\psi\\
&Inclusive(weak)\, disjunction  &\qquad \varphi\vee\psi &\qquad \mathrm{A}\varphi\psi\\
&Exclusive(strong)\, disjunction   &\qquad \varphi\not\equiv\psi &\qquad \mathrm{J}\varphi\psi\\
&Material\,implication  &\qquad \varphi\rightarrow\psi &\qquad \mathrm{C}\varphi\psi\\
&Material\,equivalence  &\qquad \varphi\leftrightarrow\psi &\qquad \mathrm{E}\varphi\psi\\
&Sheffer\,stroke(non\!-\!conjunction) &\qquad \varphi\mid\psi &\qquad \mathrm{D}\varphi\psi\\
&Peirce's\,arrow(non\!-\!disjunction) &\qquad \varphi\downarrow\psi &\qquad \mathrm{X}\varphi\psi\\
&Possibility  &\qquad \Diamond\varphi &\qquad \mathrm{M}\varphi\\
&Necessity  &\qquad \Box\varphi &\qquad \mathrm{L}\varphi\\
&Universal\,quantifier  &\qquad \forall x\varphi &\qquad \mathrm{\Pi}x\varphi\\
&Existential\,quantifier &\qquad \exists x\varphi &\qquad \mathrm{\Sigma}x\varphi\\
\end{alignat*}
So, we can rewrite the formula in question as
$$\mathrm{AKCC}ab\mathrm{C}cd\mathrm{N}ac$$
In case that the purpose is merely a notational conversion, it might be more practical to begin from the innermost parentheses: $((((\mathrm{C}ab)\Rightarrow(C\Rightarrow D))\land(\lnot A))\lor C)$ and so on.
(There may be variations in symbols; some may prefer $\not\leftrightarrow$ instead of $\not\equiv$, etc.).
