
theorem Th50:
  for A1,A2,B1,B2,C1,C2 being category, F1 being Functor of A1,B1,
      F2 being Functor of A2,B2, G1 being Functor of B1,C1,
      G2 being Functor of B2,C2
  st F1 is covariant & G1 is covariant & F2 is covariant & G2 is covariant
  holds (G1 [x] G2)(*)(F1 [x] F2) = G1(*)F1 [x] G2(*)F2
  proof
    let A1,A2,B1,B2,C1,C2 be category;
    let F1 be Functor of A1,B1;
    let F2 be Functor of A2,B2;
    let G1 be Functor of B1,C1;
    let G2 be Functor of B2,C2;
    assume
A1: F1 is covariant;
    assume
A2: G1 is covariant;
    assume
A3: F2 is covariant;
    assume
A4: G2 is covariant;
A5: F1 [x] F2 is covariant by A1,A3,Def22;
A6: G1 [x] G2 is covariant by A2,A4,Def22;
A7: G1(*)F1 is covariant by A1,A2,CAT_6:35;
A8: G2(*)F2 is covariant by A3,A4,CAT_6:35;
A9: (G1 [x] G2)(*)(F1 [x] F2) is covariant by A5,A6,CAT_6:35;
A10: (G1(*)F1)(*)pr1(A1,A2) = G1(*)(F1(*)pr1(A1,A2)) by A1,A2,CAT_7:10
    .= G1(*)(pr1(B1,B2)(*)(F1 [x] F2)) by A1,A3,Def22
    .= (G1(*)pr1(B1,B2))(*)(F1 [x] F2) by A5,A2,CAT_7:10
    .= (pr1(C1,C2)(*)(G1 [x] G2))(*)(F1 [x] F2) by A2,A4,Def22
    .= pr1(C1,C2)(*)((G1 [x] G2)(*)(F1 [x] F2)) by A5,A6,CAT_7:10;
    (G2(*)F2)(*)pr2(A1,A2) = G2(*)(F2(*)pr2(A1,A2)) by A3,A4,CAT_7:10
    .= G2(*)(pr2(B1,B2)(*)(F1 [x] F2)) by A1,A3,Def22
    .= (G2(*)pr2(B1,B2))(*)(F1 [x] F2) by A5,A4,CAT_7:10
    .= (pr2(C1,C2)(*)(G1 [x] G2))(*)(F1 [x] F2) by A2,A4,Def22
    .= pr2(C1,C2)(*)((G1 [x] G2)(*)(F1 [x] F2)) by A5,A6,CAT_7:10;
    hence thesis by A7,A10,A8,A9,Def22;
  end;
