theorem
  F|=A implies F|='G' A
 proof
  assume A1: F|=A;
  let M;
  assume A2: M|=F;
  let n be Element of NAT;
  for i be Element of NAT holds(SAT M).[n+i,A]=1 by Def12,A1,A2;
  hence (SAT M).[n,'G' A]=1 by Th10;
 end;
