theorem
  H is biconditional implies the_left_side_of (H/(x,y)) = (
the_left_side_of H)/(x,y) & the_right_side_of (H/(x,y)) = (the_right_side_of H)
  /(x,y)
proof
  assume H is biconditional;
  then H = (the_left_side_of H) <=> (the_right_side_of H) & H/(x,y) = (
  the_left_side_of (H/(x,y))) <=> (the_right_side_of (H/(x,y))) by Th176,
ZF_LANG:49;
  hence thesis by Th163;
end;
