theorem  :: (2.12)
  (a => b) = (Top L) & (b => c) = (Top L) implies (a => c) = (Top L)
  proof
    assume
A1: (a => b) = (Top L) & (b => c) = (Top L);
A2: a < b by A1;
A3: b < c by A1;
    a < c by Def3,A2,A3;
    hence thesis;
  end;
