How to prove that Ambiguity is still present in Resolved Production of Dangling Else Problem? $\textbf{stmt} \to$ $ \textbf{if} $expr$ \textbf{then}$ stmt
                           $\mid $ $\textbf{if}$ expr $ \textbf{then}$ stmt$ \textbf{else}$ stmt
                           $\mid \textbf{other} $

This grammar is called as Dangling Else Problem . However the ambiguity is resolved by parser as described as follows


  In all programming languages with conditional statements of this form, the first parse tree is preferred. 
  The general rule is, "Match each else with the  closest unmatched then"


It is rarely built into productions 

$\textbf{stmt} \to$ $ \textbf{matched_stmt}$
                          $\mid $ $\textbf{open_stmt}$

$\textbf{matched_stmt} \to$ $ \textbf{if} $expr$ \textbf{then}$ matched_stmt $ \textbf{else}$ matched_stmt
                          $\mid $ $\textbf{other}$

$\textbf{open_stmt} \to$ $ \textbf{if} $expr$ \textbf{then}$ stmt
                           $\mid $ $\textbf{if}$ expr $ \textbf{then}$ matched_stmt$ \textbf{else}$ open_stmt

However this grammar is also ambiguous . Moreover no grammar is there that could eliminate ambiguity problem of Dangling else problem . Hence we can say that dangling else problem in also inherently ambiguous  

Source : Compilers: Principles, Techniques, &amp; Tools -Aho &amp; Ullman

My question is How to prove that  Ambiguity is still present  in this resolved Production of Dangling Else Problem   ? What I know is that if two different parse trees are possible then the grammar in ambiguous . Does any such parse trees possible for the resolved grammar ? If so could u please give an example parse tree for the same
A grammar $G$ for a language $L$ is ambiguous if there is a string $w \in L$ which has two different parse trees (with respect to $G$). Hence in order to show that ambiguity is still present, all you have to do is to come up with a string generated by your grammar that has two different parse trees.