theorem Th6:
  F |=0 'G' A implies F |=0 A
  proof
    assume Z1: F |=0 'G' A;
    let M;
    assume M |=0 F;then
    M |=0 'G' A by Z1;then
    (SAT M).[0+0,A]=1 by LTLAXIO1:10;
    hence (SAT M).[0,A]=1;
  end;
