theorem onecon:
  F is consistent & not F \/ {A} is consistent implies
    F \/ {'not' A} is consistent
  proof
    assume A1: F is consistent & not F \/ {A} is consistent;
    assume not F \/ {'not' A} is consistent;then
    A2: F |- ('not'A) => A by ded,conco;
    F |- (('not' A) => A) => A by naa;then
    A6: F |- A by A2,th43;
    F |- 'not' A by A1,conco,ded;
    hence contradiction by A6,A1;
  end;
