theorem Th5:
  F |=0 'G' A implies F |=0 'G' 'X' A
 proof
  assume A1: F |=0 'G' A;
  let M;
  assume M |=0 F;then
A3: M |=0 'G' A by A1;
  now
    let i be Element of NAT;
    (SAT M).[0+(i+1),A] = 1 by LTLAXIO1:10,A3;
    hence (SAT M).[0+i,'X' A] = 1 by LTLAXIO1:9;
  end;
  hence M |=0 'G' 'X' A by LTLAXIO1:10;
 end;
