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> require"lpeg"
> 
> unaryOp = "!"        --unary Operator
> binaryOp = "&&"    --binary Operator
> expression = lpeg.P{
>     "Exp";            --Rule 1
>     Exp = lpeg.V"uExp" + lpeg.V"bExp",
>     uExp = lpeg.V"Exp" * unaryOp,
>     bExp = lpeg.V"Exp" * binaryOp * lpeg.V"Exp",
> }
> 
> This (simplified beyond sensefulness) piece of code fails: "rule
> 'bExp' is left recursive" says the interpreter (but uExp works fine).
> But wheres the problem? If it wasnt recursive, i wouldn't use a
> grammar table...  Am I doing something wrong? Or is this the reason
> for the "0.9" version number of the LPEG- lib?

This has nothing to do with 0.9. Like any other top-down parser, LPEG
cannot handle left recursion. You still can (and must) use grammars
for recursion, only left recursion is problematic. In your case,
a bExp may start with an Exp that may start with a bExp... Your grammar
is also ambiguous, so practically any parser tool will not accept it.

The reference manual for LPEG has an example that is exactly a grammar
for expressions. You may check it for some ideas.

-- Roberto