theorem Th15:
  (SAT M).[n,('not'('X' B))=>('X'('not' B))]=1
 proof
  A1: (SAT M).[n,('not'('X' B))]=0 or(SAT M).[n,('not'('X' B))]=1 by
XBOOLEAN:def 3;
  thus(SAT M).[n,('not'('X' B))=>('X'('not' B))]=(SAT M).[n,('not'('X' B))]=>(
SAT M).[n,('X'('not' B))] by Def11
   .=1 by A1,Th14;
 end;
