How to prove formally that grammar isn't LR(1) I want to prove that grammar
$$
\begin{cases}
S'\rightarrow S\\
S\rightarrow aSb ~|~ A\\
A\rightarrow bA~|~b
\end{cases}
$$
isn't $LR(1)$. I've constructed parser table and got Shift-Reduce conflict.

I want to prove that without parser table, using another $LR(1)$ definition.

Here's definition:
Grammar is $LR(1)$, if from


$S' \Rightarrow^*_r uAw \Rightarrow_r uvw$
$S' \Rightarrow^*_r zBx \Rightarrow_r uvy$
$FIRST(w) = FIRST(y)$


$\Rightarrow uAy=zBx.$

So how can prove that?
$$S'\Rightarrow^*\underbrace{ab}_uA\underbrace{b}_w\Rightarrow \underbrace{ab}_u\underbrace{b}_v\underbrace{b}_w$$

$$S'\Rightarrow^*\underbrace{abbbb}_zA\underbrace{b}_x\Rightarrow \underbrace{ab}_u\underbrace{b}_v\underbrace{bbbb}_y$$

$$FIRST(w)=FIRST(y)=b$$
But:
$$abAbbbb=uAy\neq zBx=abbbbAb$$