It's not possible to do this for regular expressions, but it is possible to do it for context free grammars. You can write a context-free grammar in Backus-Naur form that recognizes all context-free grammars in Backus-Naur form:

   <syntax>         ::= <rule> | <rule> <syntax>
   <rule>           ::= <opt-whitespace> "<" <rule-name> ">" <opt-whitespace> "::=" <opt-whitespace> <expression> <line-end>
   <opt-whitespace> ::= " " <opt-whitespace> | ""
   <expression>     ::= <list> | <list> <opt-whitespace> "|" <opt-whitespace> <expression>
   <line-end>       ::= <opt-whitespace> <EOL> | <line-end> <line-end>
   <list>           ::= <term> | <term> <opt-whitespace> <list>
   <term>           ::= <literal> | "<" <rule-name> ">"
   <literal>        ::= '"' <text1> '"' | "'" <text2> "'"
   <text1>          ::= "" | <character1> <text1>
   <text2>          ::= '' | <character2> <text2>
   <character>      ::= <letter> | <digit> | <symbol>
   <letter>         ::= "A" | "B" | "C" | "D" | "E" | "F" | "G" | "H" | "I" | "J" | "K" | "L" | "M" | "N" | "O" | "P" | "Q" | "R" | "S" | "T" | "U" | "V" | "W" | "X" | "Y" | "Z" | "a" | "b" | "c" | "d" | "e" | "f" | "g" | "h" | "i" | "j" | "k" | "l" | "m" | "n" | "o" | "p" | "q" | "r" | "s" | "t" | "u" | "v" | "w" | "x" | "y" | "z"
   <digit>          ::= "0" | "1" | "2" | "3" | "4" | "5" | "6" | "7" | "8" | "9"
   <symbol>         ::=  "|" | " " | "!" | "#" | "$" | "%" | "&" | "(" | ")" | "*" | "+" | "," | "-" | "." | "/" | ":" | ";" | ">" | "=" | "<" | "?" | "@" | "[" | "\" | "]" | "^" | "_" | "`" | "{" | "}" | "~"
   <character1>     ::= <character> | "'"
   <character2>     ::= <character> | '"'
   <rule-name>      ::= <letter> | <rule-name> <rule-char>
   <rule-char>      ::= <letter> | <digit> | "-"

I got this from Wikipedia: https://en.wikipedia.org/wiki/Backus%E2%80%93Naur_form#Furth...