How do I eliminate left recursion in discrete math I am working on a problem and the grammar is below.
$A\rightarrow AA$*  
 $\rightarrow AA+$ 
 $\rightarrow a$ 
The formula for elimination is 
$A\rightarrow Aα|β$ 
$A\rightarrow βA'$
$A'\rightarrow αA'$ 
I just don't know how to use that formula if I have two AA instead of just one A 

$A->aA'$
$A'\rightarrow A+A'$ 
  $\rightarrow A*A'$ 
  $\rightarrow a$ 
However, I still have one $A$ in there. How do I get rid of that?
Thank you
 A: I hope it can help you.

Elimination of Left-Recursion
For each rule which contains a left-recursive option,
$$A \to A\alpha|\beta$$ 
introduce a new nonterminal A' and rewrite the
  rule as
\begin{align}
 &A \to \beta A' \\ 
 &A' \to  \alpha A'|\epsilon
\end{align}
there may be more than one left-recursive part on the right-hand side.
  The general rule is to replace:
$$A \to A  \alpha_1 | \alpha_2|...| \alpha_n|\beta_1|\beta_2| ...
 |\beta_m $$  
by
   \begin{align}
 &A \to \beta_1  A' | \beta_2  A'| ...|   \beta_m A' \\
 &A' \to \epsilon | \alpha_1 A' |  \alpha_2A' | ...|  \alpha_n A'
\end{align}


$ A\to AA*|AA+|a $
we can write it as:
$A\to AB|AC|a \\
B \to A* \\
C \to A+ $



*

*$A\to AB|a :\\
  \,\,\,A\to aA'\\
 \,\,\,\,A' \to A*A'|\epsilon$

*$A \to AC|a :\\
 \,\,\,A \to aA'\\
 \,\,\,\,A' \to A+A'|\epsilon$
combine those two:
$A \to aA'\\ 
A' \to A+A'|A*A'|\epsilon$
A: As you mention, the formula for removing left-recursion (and replacing it with right-recursion) involves replacing all instances of this rule:
\begin{gather}
A \rightarrow A\alpha\ |\ \beta\\
\end{gather}
with this rule:
\begin{gather}
A \rightarrow \beta A'\\
A' \rightarrow \alpha A'\ |\ \epsilon
\end{gather}
Your grammar is given by the following rules:
\begin{gather}
A \rightarrow AA*\ |\ AA+\ |\ a\\
\end{gather}
Here, we must eliminate left-recursion in two places, using the above rule for each case:


*

*$A \rightarrow AA*\ |\ a$ (here, $\alpha = A*$ and $\beta=a$)
\begin{gather}
A \rightarrow a A'\\
A' \rightarrow A*A'\ |\ \epsilon
\end{gather}

*$A \rightarrow AA+\ |\ a$ (here, $\alpha = A+$ and $\beta=a$).
\begin{gather}
A \rightarrow a A'\\
A' \rightarrow A+A'\ |\ \epsilon
\end{gather}


Combining these, we obtain:
\begin{gather}
A \rightarrow a A'\\
A' \rightarrow A+A'\ |\ A*A'\ |\ \epsilon
\end{gather}
A: Is this the grammar you've come up with? 

A->aA'
  A'->A+A'
  A'->A*A'
  A'->a 

If so, this grammar does not have any left-recursion, so it looks like you're already finished. 
(However, shouldn't the last rule produce the empty string instead of "a"?) 
