reserve i, j, k, c, m, n for Nat,
  a, x, y, z, X, Y for set,
  D, E for non empty set,
  R for Relation,
  f, g for Function,
  p, q for FinSequence;

theorem Th3:
  for i being Nat holds Del(p+*(i,x),i) = Del(p,i)
proof let i be Nat;
  set f = p;
  per cases;
  suppose
A1: i in dom f;
    then
A2: i <= len f by FINSEQ_3:25;
    1 <= i by A1,FINSEQ_3:25;
    then consider j being Nat such that
A3: len f = j+1 by A2,NAT_1:6;
    reconsider j as Element of NAT by ORDINAL1:def 12;
A4: dom (f+*(i,x)) = dom f by FUNCT_7:30;
    then
A5: len (f+*(i,x)) = len f by FINSEQ_3:29;
    then len Del(f+*(i,x),i) = j by A1,A3,A4,FINSEQ_3:109;
    then
A6: dom Del(f+*(i,x),i) = Seg j by FINSEQ_1:def 3;
    now
      thus len Del(f+*(i,x),i) = j by A1,A3,A4,A5,FINSEQ_3:109;
      thus len Del(f,i) = j by A1,A3,FINSEQ_3:109;
      let a be Nat;
      assume a in dom Del(f+*(i,x),i);
      then
A7:   a <= j by A6,FINSEQ_1:1;
      per cases;
      suppose
A8:    a < i;
        hence Del(f+*(i,x),i).a = (f+*(i,x)).a by FINSEQ_3:110
          .= f.a by A8,FUNCT_7:32
          .= Del(f,i).a by A8,FINSEQ_3:110;
      end;
      suppose
A9:    i <= a;
        then
A10:    i < a+1 by NAT_1:13;
        thus Del(f+*(i,x),i).a = (f+*(i,x)).(a+1) by A1,A3,A4,A5,A7,A9,
FINSEQ_3:111
          .= f.(a+1) by A10,FUNCT_7:32
          .= Del(f,i).a by A1,A3,A7,A9,FINSEQ_3:111;
      end;
    end;
    hence thesis by FINSEQ_2:9;
  end;
  suppose
    not i in dom f;
    hence thesis by FUNCT_7:def 3;
  end;
end;
