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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

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

$\Rightarrow uAy=zBx.$

So how can prove that?

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$$S'\Rightarrow^*\underbrace{ab}_uA\underbrace{b}_w\Rightarrow \underbrace{ab}_u\underbrace{b}_v\underbrace{b}_w$$

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

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

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  • $\begingroup$ can u prove above definition?? $\endgroup$ – T.J. Nov 4 '13 at 5:09

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