theorem
  M |= H1 => All(x,H2) implies M |= H1 => H2
proof
  assume
A1: for v holds M,v |= H1 => All(x,H2);
  let v;
A2: M,v |= H1 => All(x,H2) by A1;
  now
    assume M,v |= H1;
    then M,v |= All(x,H2) by A2,ZF_MODEL:18;
    then M,v/(x,v.x) |= H2 by Th71;
    hence M,v |= H2 by FUNCT_7:35;
  end;
  hence thesis by ZF_MODEL:18;
end;
