
theorem Th50:
for f,g be sequence of ExtREAL, i,j be Nat st
 f is nonnegative & f.i = g.j & f.j = g.i
 & (for n be Nat st n <> i & n <> j holds f.n = g.n)
 holds for n be Nat st n >= i & n >= j holds (Ser f).n = (Ser g).n
proof
    let f,g be sequence of ExtREAL, i,j be Nat;
    assume that
A1:  f is nonnegative and
A2:  f.i = g.j and
A3:  f.j = g.i and
A4:  for n be Nat st n <> i & n <> j holds f.n = g.n;

    let n be Nat;
    assume
A5:  n >= i & n >= j;

    defpred P[Nat] means $1 >= i & $1 >= j implies (Ser f).$1 = (Ser g).$1;

    now assume 0 >= i & 0 >= j; then
     i = 0 & j = 0; then
     (Ser f).0 = g.0 by A2,SUPINF_2:def 11;
     hence (Ser f).0 = (Ser g).0 by SUPINF_2:def 11;
    end; then
A6: P[0];

A7: for k be Nat st P[k] holds P[k+1]
    proof
     let k be Nat;
     assume
A8:   P[k];
     now assume
A9:   k+1 >= i & k+1 >= j;

      per cases;
      suppose k < i & k < j; then
       k+1 <= i & k+1 <= j by NAT_1:13; then
       k+1 = i & k+1 = j by A9,XXREAL_0:1;
       hence (Ser f).(k+1) = (Ser g).(k+1) by A1,A3,A4,Th49;
      end;
      suppose
A10:   k >= i & k < j; then
       k+1 <= j by NAT_1:13; then
A11:   k+1 = j by A9,XXREAL_0:1;
       for n be Nat st n <> j & n <> i holds f.n = g.n by A4;
       hence (Ser f).(k+1) = (Ser g).(k+1)
         by A1,A2,A3,A11,A10,NAT_1:12,Th49;
      end;
      suppose
A12:   k < i & k >= j; then
       k+1 <= i by NAT_1:13; then
       k+1 = i by A9,XXREAL_0:1;
       hence (Ser f).(k+1) = (Ser g).(k+1)
         by A1,A2,A3,A4,A12,NAT_1:12,Th49;
      end;
      suppose
A13:   k >= i & k >= j; then
A14:   k+1 > i & k+1 > j by NAT_1:13;
       (Ser f).(k+1) = (Ser f).k + f.(k+1) by SUPINF_2:def 11
        .= (Ser g).k + g.(k+1) by A4,A8,A13,A14;
       hence (Ser f).(k+1) = (Ser g).(k+1) by SUPINF_2:def 11;
      end;
     end;
     hence P[k+1];
    end;

    for k be Nat holds P[k] from NAT_1:sch 2(A6,A7);
    hence (Ser f).n = (Ser g).n by A5;
end;
