Question: Show the string $$\forall v_1\forall v_2(Pv_1\rightarrow Pv_2\rightarrow\equiv fv_1v_2c)$$ is not an $\mathcal{L}$-formula
Answer: Assume for contradiction that $\forall v_1\forall v_2(Pv_1\rightarrow Pv_2\rightarrow\equiv fv_1v_2c)$ is an $\mathcal{L}$-formula
$(1)$ By unique readability twice, we have that $(Pv_1\rightarrow Pv_2\rightarrow\equiv fv_1v_2c)$ is an $\mathcal{L}$-formula
$...$ etc
Contention: How does the unique readability theorem show it is an $\mathcal{L}$-formula?
Does it deduce it from eliminating the $\forall v_1$ and $\forall v_2$ before the brackets,
or is it from comprising what is in the brackets to make up the $\mathcal{L}$-formula
(I am suspecting the former, but I would like some clarification)
I can see that $(Pv_1)$, $(Pv_2)$ and $(\equiv fv_1v_2c)\in AtFml_\mathcal{L}$
I would just like some clarrification on how the unique readability theorem works for $\mathcal{L}$-formulas
In a language/grammar with unique readability, a well formed formula has only one unique parse tree.
In many languages the infix notation $\varphi_1 \rightarrow \varphi_2 \rightarrow \varphi_3$ has unique readability by convention, but apparently in your language this is not the case. It is easy to see the ambiguity by converting to prefix notation. One parsing is $\rightarrow( \varphi_1, \rightarrow(\varphi_2, \varphi_3)) $, while the other is $\rightarrow( \rightarrow(\varphi_1, \varphi_2), \varphi_3)$.
Since you assumed that your string was well formed, it should have unique readability. Since we have shown that it does not, we have found a contradiction and your assumption must not be correct.