theorem Th10:
  Terminals EmptyGrammar a = {}
proof
  set E = EmptyGrammar a;
  set b = the Element of Terminals E;
  the Rules of E = {[a,{}]} by Def7;
  then
A1: [a,{}] in the Rules of E by TARSKI:def 1;
  assume not thesis;
  then b in Terminals E;
  then consider s being Symbol of E such that
  b = s and
A2: not ex n being FinSequence st s ==> n;
  the carrier of E = {a} by Def7;
  then s = a by TARSKI:def 1;
  then s ==> <*>the carrier of E by A1;
  hence contradiction by A2;
end;
