reserve A,B,C,D for Category,
  F for Functor of A,B,
  G for Functor of B,C;
reserve o,m for set;
reserve F,F1,F2,F3 for Functor of A,B,
  G,G1,G2,G3 for Functor of B,C,
  H,H1,H2 for Functor of C,D,
  s for natural_transformation of F1,F2,
  s9 for natural_transformation of F2,F3,
  t for natural_transformation of G1,G2,
  t9 for natural_transformation of G2,G3,
  u for natural_transformation of H1,H2;

theorem
  A,B are_equivalent & B,C are_equivalent implies for F being
Equivalence of A,B, G being Equivalence of B,C holds G*F is Equivalence of A,C
proof
  assume that
A1: A,B are_equivalent and
A2: B,C are_equivalent;
  let F be Equivalence of A,B, G be Equivalence of B,C;
  thus A,C are_equivalent by A1,A2,Th44;
  consider F9 being Functor of B,A such that
A3: F9*F ~= id A and
A4: F*F9 ~= id B by A1,Def11;
  (G*F)*F9 = G*(F*F9) by RELAT_1:36;
  then
A5: (G*F)*F9 ~= G by A4,Th42;
  consider G9 being Functor of C,B such that
A6: G9*G ~= id B and
A7: G*G9 ~= id C by A2,Def11;
  take F9*G9;
  (F9*G9)*G = F9*(G9*G) by RELAT_1:36;
  then
A8: (F9*G9)*G ~= F9 by A6,Th42;
  (F9*G9)*(G*F) = ((F9*G9)*G)*F by RELAT_1:36;
  then (F9*G9)*(G*F) ~= F9*F by A8,Th41;
  hence (F9*G9)*(G*F) ~= id A by A3,NATTRA_1:29;
  (G*F)*(F9*G9) = ((G*F)*F9)*G9 by RELAT_1:36;
  then (G*F)*(F9*G9) ~= G*G9 by A5,Th41;
  hence thesis by A7,NATTRA_1:29;
end;
