
theorem Th44:
  for f1,f2 be without-infty Function of [:NAT,NAT:],ExtREAL holds
   Partial_Sums_in_cod2(f1+f2)
    = Partial_Sums_in_cod2 f1 + Partial_Sums_in_cod2 f2 &
   Partial_Sums_in_cod1(f1+f2)
    = Partial_Sums_in_cod1 f1 + Partial_Sums_in_cod1 f2
proof
   let f1,f2 be without-infty Function of [:NAT,NAT:],ExtREAL;
   set CS1 = Partial_Sums_in_cod2 f1;
   set CS2 = Partial_Sums_in_cod2 f2;
   set CS12 = Partial_Sums_in_cod2(f1+f2);
   set RS1 = Partial_Sums_in_cod1 f1;
   set RS2 = Partial_Sums_in_cod1 f2;
   set RS12 = Partial_Sums_in_cod1(f1+f2);
   now let n be Element of NAT, m be Element of NAT;
    defpred C[Nat] means CS12.(n,$1) = CS1.(n,$1) + CS2.(n,$1);
    CS12.(n,0) = (f1+f2).(n,0) by DefCSM
     .= f1.(n,0) + f2.(n,0) by Th11
     .= CS1.(n,0) + f2.(n,0) by DefCSM; then
a1: C[0] by DefCSM;
a2: for k be Nat st C[k] holds C[k+1]
    proof
     let k be Nat;
     assume a3: C[k];
X1:  CS1.(n,k) <> -infty & CS2.(n,k) <> -infty & f1.(n,k+1) <> -infty
   & f2.(n,k+1) <> -infty & (f1+f2).(n,k+1) <> -infty
         by MESFUNC5:def 5; then
X2:  CS2.(n,k) + f2.(n,k+1) <> -infty by XXREAL_3:17;
     CS12.(n,k+1) = CS12.(n,k) + (f1+f2).(n,k+1) by DefCSM
      .= CS1.(n,k) + ( (f1+f2).(n,k+1) + CS2.(n,k) ) by a3,X1,XXREAL_3:29
      .= CS1.(n,k) + ( f1.(n,k+1) + f2.(n,k+1) + CS2.(n,k) ) by Th11
      .= CS1.(n,k) + ( f1.(n,k+1) + ( CS2.(n,k) + f2.(n,k+1) ) )
           by X1,XXREAL_3:29
      .= CS1.(n,k) + f1.(n,k+1) + ( CS2.(n,k) + f2.(n,k+1) )
           by X1,X2,XXREAL_3:29
      .= CS1.(n,k+1) + ( CS2.(n,k) + f2.(n,k+1) ) by DefCSM;
     hence C[k+1] by DefCSM;
    end;
    for k be Nat holds C[k] from NAT_1:sch 2(a1,a2); then
    C[m];
    hence CS12.(n,m) = (CS1+CS2).(n,m) by Th11;
   end;
   hence CS12 = CS1 + CS2 by BINOP_1:2;
   now let n,m be Element of NAT;
    defpred R[Nat] means RS12.($1,m) = RS1.($1,m) + RS2.($1,m);
    RS12.(0,m) = (f1+f2).(0,m) by DefRSM
     .= f1.(0,m) + f2.(0,m) by Th11
     .= RS1.(0,m) + f2.(0,m) by DefRSM; then
a4: R[0] by DefRSM;
a5: for k be Nat st R[k] holds R[k+1]
    proof
     let k be Nat;
     assume a6: R[k];
X3:  RS1.(k,m) <> -infty & RS2.(k,m) <> -infty & f1.(k+1,m) <> -infty
   & f2.(k+1,m) <> -infty & (f1+f2).(k+1,m) <> -infty
         by MESFUNC5:def 5; then
X4:  RS2.(k,m) + f2.(k+1,m) <> -infty by XXREAL_3:17;
     RS12.(k+1,m) = RS12.(k,m) + (f1+f2).(k+1,m) by DefRSM
      .= RS1.(k,m) + ( (f1+f2).(k+1,m) + RS2.(k,m) ) by a6,X3,XXREAL_3:29
      .= RS1.(k,m) + ( f1.(k+1,m) + f2.(k+1,m) + RS2.(k,m) ) by Th11
      .= RS1.(k,m) + ( f1.(k+1,m) + ( RS2.(k,m) + f2.(k+1,m) ) )
           by X3,XXREAL_3:29
      .= RS1.(k,m) + f1.(k+1,m) + ( RS2.(k,m) + f2.(k+1,m) )
           by X3,X4,XXREAL_3:29
      .= RS1.(k+1,m) + (RS2.(k,m) + f2.(k+1,m)) by DefRSM;
     hence R[k+1] by DefRSM;
    end;
    for k be Nat holds R[k] from NAT_1:sch 2(a4,a5); then
    R[n];
    hence RS12.(n,m) = (RS1+RS2).(n,m) by Th11;
   end;
   hence RS12 = RS1 + RS2 by BINOP_1:2;
end;
