reserve k,n,m,l,p for Nat;
reserve n0,m0 for non zero Nat;
reserve f for FinSequence;
reserve x,X,Y for set;

theorem Th7:
  f is one-to-one & n in dom f implies not f.n in rng Del(f,n)
proof
  assume
A1: f is one-to-one;
  reconsider n9=n as Element of NAT by ORDINAL1:def 12;
A2: dom Del(f,n9) c= dom f by WSIERP_1:39;
  assume
A3: n in dom f;
  then consider m be Nat such that
A4: len f = m + 1 and
A5: len Del(f,n) = m by FINSEQ_3:104;
  assume f.n in rng Del(f,n);
  then consider j be object such that
A6: j in dom Del(f,n) and
A7: f.n = Del(f,n).j by FUNCT_1:def 3;
A8: j in Seg m by A5,A6,FINSEQ_1:def 3;
  reconsider j as Element of NAT by A6;
  per cases;
  suppose
A9: j<n9;
    then f.n = f.j by A7,FINSEQ_3:110;
    hence contradiction by A1,A3,A6,A2,A9,FUNCT_1:def 4;
  end;
  suppose
A10: j>=n9;
A11: j+1 >= 0+1 by XREAL_1:6;
A12: j <= m by A8,FINSEQ_1:1;
    then j+1 <= m+1 by XREAL_1:6;
    then j+1 in Seg(m+1) by A11;
    then
A13: j+1 in dom f by A4,FINSEQ_1:def 3;
A14: j+1 <> n by A10,NAT_1:13;
    f.n = f.(j+1) by A3,A4,A7,A10,A12,FINSEQ_3:111;
    hence contradiction by A1,A3,A13,A14,FUNCT_1:def 4;
  end;
end;
