reserve A,B,C,D for Category,
  F for Functor of A,B,
  G for Functor of B,C;
reserve o,m for set;

theorem
  A ~= B & C ~= D implies [:A,C:] ~= [:B,D:]
proof
  given F being Functor of A,B such that
A1: F is isomorphic;
A2: F is one-to-one by A1;
  given G being Functor of C,D such that
A3: G is isomorphic;
  take [:F,G:];
  G is one-to-one by A3;
  hence [:F,G:] is one-to-one by A2;
  thus rng [:F,G:] = [:rng F, rng G:] by FUNCT_3:67
    .= [: the carrier' of B, rng G:] by A1
    .= the carrier' of [:B,D:] by A3;
end;
