theorem
  F is_proper_subformula_of 'not' H implies F is_subformula_of H
proof
  assume
A1: F is_proper_subformula_of 'not' H;
A2: 'not' H is negative;
  then the_argument_of ('not' H) = H by Def18;
  hence thesis by A1,A2,Th37;
end;
