Let $a,b,c$ denote elements of some field, say, the real numbers. Consider the statement
If $ac=bc$ and if $c\neq 0$ then $a=b\qquad (\textrm{I})$
Denoting $ac=bc$ by $p$, $a=b$ by $q$ and $c=0$ by $r$, a straightforward formalisation of the above statement is this:
$p\wedge(\neg r)\to q\qquad (1)$
Consider
$\neg p\to (\neg q\vee\neg r)\qquad (2)$
The second statement $(2)$ may be translated to
If $ac\neq bc$ then $a\neq b$ or $c\neq 0\qquad\textrm{(II)}$
and let's examine the truth-table of both formal statements $(1),(2)$:
\begin{array}{|c|c|c|c|c|} \hline p& q & r & p\wedge(\neg r)\to q&\neg p\to (\neg q\vee\neg r)\\ \hline T& T& T&T&T\\ \hline T& T& F&T&T\\ \hline T& F& T&T&T\\ \hline T&F&F&\color{red}F&\color{red}T\\ \hline F&T&T&\color{red}T&\color{red}F\\ \hline F&T&F&T&T\\ \hline F&F&T&T&T\\ \hline F&F&F&T&T\\ \hline \end{array}
Clearly they are not tautologically equivalent. But in the given context, where $a,b,c$ are real numbers (or elements of any field for that matter), the fourth and fifth lines of the truth table are not possible, because the fourth line means that $ac=bc$ and $a\neq b$ and $c\neq 0$, which is impossible, and the fifth line means that $ac\neq bc$ and $a=b$ and $c=0$, which is also impossible.
Question: Which of the formal statements $(1),(2)$ correctly formalises the statement $\textrm{(I)}$? Does the fact that the fourth and fifth lines are impossible imply that both statements $(1),(2)$ are correct formalisations?