theorem Th24:
 F|=A & F|=A=>B implies F|=B
 proof
  assume that
   A1: F|=A and
   A2: F|=A=>B;
  let M;
  assume A3: M|=F;
  let n be Element of NAT;
 (SAT M).[n,A=>B]=1 by Def12,A2,A3;
  then A4: (SAT M).[n,A]=>(SAT M).[n,B]=1 by Def11;
  (SAT M).[n,A]=1 by Def12,A1,A3;
  hence (SAT M).[n,B]=1 by A4;
 end;
