theorem
  not x in Free H1 & M |= H1 => H2 implies M |= H1 => All(x,H2)
proof
  assume that
A1: not x in Free H1 and
A2: for v holds M,v |= H1 => H2;
  let v;
  M |= All(x,H1 => H2) => (H1 => All(x,H2)) by A1,Th128;
  then
A3: M,v |= All(x,H1 => H2) => (H1 => All(x,H2));
  for m holds M,v/(x,m) |= H1 => H2 by A2;
  then M,v |= All(x,H1 => H2) by Th71;
  hence thesis by A3,ZF_MODEL:18;
end;
