theorem
  a delta (b"/\"c) = (a delta b)"/\"(a delta c) & (b"/\"c) delta a = (b
  delta a)"/\"(c delta a)
proof
  thus a delta (b"/\"c) = Bottom (Bottom a [*] (Bottom b"\/"Bottom c)) by Th26
    .= Bottom ((Bottom a [*] Bottom b)"\/"(Bottom a [*] Bottom c)) by Th6
    .= (a delta b)"/\"(a delta c) by Th26;
  thus (b"/\"c) delta a = Bottom ((Bottom b"\/"Bottom c) [*] Bottom a) by Th26
    .= Bottom ((Bottom b [*] Bottom a)"\/"(Bottom c [*] Bottom a)) by Th6
    .= (b delta a)"/\"(c delta a) by Th26;
end;
