Problem with LPEG- grammar

[Wolfgang Pupp <wolfgang_pupp@...>]

Trying to realise a parser for boolean expressions (its, admittedly, my first "project" with LPEG), 
i stumbled across a problem i couldn't solve:

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?

By the way, the LPEG- library is just great- I really enjoyed to use it (std. regular expressions with Lua? never again!) till this problem arised...

Wolfgang

PS: Thanks in advance for your help (and for Lua and LPEG)!