theorem Th14:
  (SAT M).[n,'not'('X' B)]=(SAT M).[n,'X'('not' B)]
 proof
  thus(SAT M).[n,'not'('X' B)]=(SAT M).[n,('X' B)]=>(SAT M).[n,TFALSUM] by
Def11
   .=(SAT M).[n+1,B]=>(SAT M).[n,TFALSUM] by Th9
   .=(SAT M).[n+1,B]=>FALSE by Def11
   .=(SAT M).[n+1,B]=>(SAT M).[n+1,TFALSUM] by Def11
   .=(SAT M).[n+1,B=>TFALSUM] by Def11
   .=(SAT M).[n,'X'('not' B)] by Th9;
 end;
