
theorem Th25:
  for A, B, C, D being category st A, B are_opposite & C, D are_opposite
  holds A, C are_equivalent implies B, D are_equivalent
proof
  let A, B, C, D be category;
  assume that
A1: A, B are_opposite and
A2: C, D are_opposite;
  given F being covariant Functor of A,C,
  G being covariant Functor of C,A such that
A3: G*F, id A are_naturally_equivalent and
A4: F*G, id C are_naturally_equivalent;
  take dF = dualizing-func(C,D)*F*dualizing-func(B,A),
  dG = dualizing-func(A,B)*G*dualizing-func(D,C);
A5: G* id C = the FunctorStr of G by FUNCTOR3:5;
A6: dualizing-func(A,B)*(id A) = dualizing-func(A,B) by FUNCTOR3:5;
A7: id C = dualizing-func(D,C)*dualizing-func(C,D) by A2,Th14;
A8: dualizing-func(A,B)*(G*F)*dualizing-func(B,A)
  = dualizing-func(A,B)*G*F*dualizing-func(B,A) by FUNCTOR0:32
    .= dualizing-func(A,B)*G*(F*dualizing-func(B,A)) by FUNCTOR0:32
    .= dualizing-func(A,B)*(G*(id C))*(F*dualizing-func(B,A)) by A5,Th3
    .= dualizing-func(A,B)*G*(id C)*(F*dualizing-func(B,A)) by FUNCTOR0:32
    .= dG*dualizing-func(C,D)*(F*dualizing-func(B,A)) by A7,FUNCTOR0:32
    .= dG*(dualizing-func(C,D)*(F*dualizing-func(B,A))) by FUNCTOR0:32
    .= dG*dF by FUNCTOR0:32;
  dualizing-func(A,B)*(id A)*dualizing-func(B,A) = id B by A1,A6,Th14;
  hence dG*dF, id B are_naturally_equivalent by A1,A3,A8,Th24;
A9: F* id A = the FunctorStr of F by FUNCTOR3:5;
A10: dualizing-func(C,D)*(id C) = dualizing-func(C,D) by FUNCTOR3:5;
A11: id A = dualizing-func(B,A)*dualizing-func(A,B) by A1,Th14;
A12: dualizing-func(C,D)*(F*G)*dualizing-func(D,C)
  = dualizing-func(C,D)*F*G*dualizing-func(D,C) by FUNCTOR0:32
    .= dualizing-func(C,D)*F*(G*dualizing-func(D,C)) by FUNCTOR0:32
    .= dualizing-func(C,D)*(F*(id A))*(G*dualizing-func(D,C)) by A9,Th3
    .= dualizing-func(C,D)*F*(id A)*(G*dualizing-func(D,C)) by FUNCTOR0:32
    .= dF*dualizing-func(A,B)*(G*dualizing-func(D,C)) by A11,FUNCTOR0:32
    .= dF*(dualizing-func(A,B)*(G*dualizing-func(D,C))) by FUNCTOR0:32
    .= dF*dG by FUNCTOR0:32;
  dualizing-func(C,D)*(id C)*dualizing-func(D,C) = id D by A2,A10,Th14;
  hence thesis by A2,A4,A12,Th24;
end;
