reserve i for Integer,
  a, b, r, s for Real;

theorem
  |[a,b]|.1 = a & |[a,b]|.2 = b
proof
  thus |[a,b]|.1 = ((1,2) --> (a,b)).1 by Th4
    .= a by FUNCT_4:63;
  thus |[a,b]|.2 = ((1,2) --> (a,b)).2 by Th4
    .= b by FUNCT_4:63;
end;
