theorem X |- p => (q '&&' ('not' q)) implies X |- 'not' p
  proof
    (p => (q '&&' ('not' q))) => ('not' p) is ctaut by Th43;then
    (p => (q '&&' ('not' q))) => ('not' p) in LTL_axioms by LTLAXIO1:def 17;
    then A1: X |- (p => (q '&&' ('not' q))) => ('not' p) by LTLAXIO1:42;
    assume X |- p => (q '&&' ('not' q));
    hence X |- 'not' p by A1,LTLAXIO1:43;
  end;
