grammar:                       left recursive grammar hack:
 S      -> Assign EOL           S  -> E$         F  -> AF'
         | Query EOL            E  -> TE'        F' -> not A
 Assign -> id eq Exp            E' -> and TE'        | eq  A
 Query  -> id qmark                 | e              | id
 Exp    -> Exp and Exp          T  -> FT'            | e
         | Exp or  Exp          T' -> or FT'      A  -> ?
         | not Exp                  | e              | 0
         | lp Exp rp                                 | 1
         | id                                        | (E)
         | true                     
         | false                    
 
 terminals:                     first sets:
 id    :: a-z                   first(S)  = (, id, 0, 1
 and   :: &                     first(E)  = (, id, 0, 1
 or    :: |                     first(E') = and
 not   :: ~                     first(T)  = (, id, 0, 1
 eq    :: =                     first(T') = or
 qmark :: ?                     first(F)  = (, id, 0, 1
 false :: 0                     first(F') = not, eq
 true  :: 1                     first(A)  = (, id, 0, 1
 lp    :: (
 rp    :: )                     follows sets:
 EOL   :: \n                    follows(S)  = $
                                follows(E)  = ), $
                                follows(E') = ), $
                                follows(T)  = ), and, or, $
                                follows(T') = ), and, or, $
                                follows(F)  = ), and, or, $
                                follows(F') = ), and, or, $
                                follows(A)  = ), and, or, $