reserve x,y for set;
reserve C,C9,D,E for non empty set;
reserve c,c9,c1,c2,c3 for Element of C;
reserve B,B9,B1,B2 for Element of Fin C;
reserve A for Element of Fin C9;
reserve d,d1,d2,d3,d4,e for Element of D;
reserve F,G for BinOp of D;
reserve u for UnOp of D;
reserve f,f9 for Function of C,D;
reserve g for Function of C9,D;
reserve H for BinOp of E;
reserve h for Function of D,E;
reserve i,j for Nat;
reserve s for Function;
reserve p,q for FinSequence of D;
reserve T1,T2 for Element of i-tuples_on D;

theorem Th9:
  F is commutative associative & F is having_a_unity & e =
the_unity_wrt F & G.(e,e) = e & (for d1,d2,d3,d4 holds F.(G.(d1,d2),G.(d3,d4))=
  G.(F.(d1,d3),F.(d2,d4))) implies G.(F$$(B,f),F$$(B,f9)) = F $$(B,G.:(f,f9))
proof
  assume that
A1: F is commutative associative & F is having_a_unity and
A2: e = the_unity_wrt F and
A3: G.(e,e) = e and
A4: for d1,d2,d3,d4 holds F.(G.(d1,d2),G.(d3,d4))= G.(F.(d1,d3),F.(d2,d4 ));
  defpred X[Element of Fin C] means G.(F$$($1,f),F$$($1,f9)) = F $$($1,G.:(f,
  f9));
A5: for B9 being Element of Fin C, b being Element of C holds X[B9] & not b
  in B9 implies X[B9 \/ {.b.}]
  proof
    let B,c such that
A6: G.(F$$(B,f),F$$(B,f9)) = F $$(B,G.:(f,f9)) and
A7: not c in B;
    set s9 = F$$(B,f9);
    set s = F$$(B,f);
    F$$(B \/ {.c.},f) = F.(s,f.c) & F$$(B \/ {.c.},f9) = F.(s9,f9.c) by A1,A7
,Th2;
    hence G.(F$$(B \/ {.c.},f),F$$(B \/ {.c.},f9)) = F.(G.(s,s9),G.(f.c,f9.c))
    by A4
      .= F.(G.(s,s9),G.:(f,f9).c) by FUNCOP_1:37
      .= F $$(B \/ {.c.},G.:(f,f9)) by A1,A6,A7,Th2;
  end;
  F$$({}.C,f) = e & F$$({}.C,f9) = e by A1,A2,SETWISEO:31;
  then
A8: X[{}.C] by A1,A2,A3,SETWISEO:31;
  for B holds X[B] from SETWISEO:sch 2(A8,A5);
  hence thesis;
end;
