theorem Th17:
  (SAT M).[n,('X'(B=>C))=>(('X' B)=>('X' C))]=1
 proof
  A1: (SAT M).[n+1,B]=0 or(SAT M).[n+1,B]=1 by XBOOLEAN:def 3;
  A2: (SAT M).[n+1,C]=0 or(SAT M).[n+1,C]=1 by XBOOLEAN:def 3;
  thus(SAT M).[n,('X'(B=>C))=>(('X' B)=>('X' C))]=(SAT M).[n,'X'(B=>C)]=>(SAT M
).[n,('X' B)=>('X' C)] by Def11
   .=(SAT M).[n+1,B=>C]=>(SAT M).[n,('X' B)=>('X' C)] by Th9
   .=(SAT M).[n+1,B=>C]=>((SAT M).[n,'X' B]=>(SAT M).[n,'X' C]) by Def11
   .=(SAT M).[n+1,B=>C]=>((SAT M).[n+1,B]=>(SAT M).[n,'X' C]) by Th9
   .=(SAT M).[n+1,B=>C]=>((SAT M).[n+1,B]=>(SAT M).[n+1,C]) by Th9
   .=1 by A1,A2,Def11;
 end;
