theorem th34:
  not F |- A implies F \/ {'not' A} is consistent
  proof
    assume Z1: not F |- A;
    assume not F \/ {'not' A} is consistent;then
A2: F |- 'not'A => A by ded,conco;
    F |- ('not' A => A) => A by naa;
    hence contradiction by Z1,A2,th43;
  end;
