theorem Th3:
  F |=0 A & F |=0 A=>B implies F |=0 B
 proof
  assume that
   A1: F |=0 A and
   A2: F |=0 A=>B;
   let M;
   assume
B3: M |=0 F;then
   M |=0 A=>B by A2;then
B6: (SAT M).[0,A] => (SAT M).[0,B] = 1 by LTLAXIO1:def 11;
   M |=0 A by A1,B3;
   hence M |=0 B by B6;
 end;
