reserve n for Nat;

theorem
  for f be FinSequence of TOP-REAL 2 st f is being_S-Seq for p be Point
  of TOP-REAL 2 st p in L~f & f.1 in L~L_Cut(f,p) holds f.1 = p
proof
  let f be FinSequence of TOP-REAL 2;
  assume
A1: f is being_S-Seq;
A2: len f in dom f by A1,FINSEQ_5:6;
  1 in dom f by A1,FINSEQ_5:6;
  then
A3: f/.1 = f.1 by PARTFUN1:def 6;
  let p be Point of TOP-REAL 2 such that
A4: p in L~f and
A5: f.1 in L~L_Cut(f,p) and
A6: f.1 <> p;
  set g = mid(f,Index(p,f)+1,len f);
A7: not f.1 in L~g by A1,A4,Th5;
  then p<>f.(Index(p,f)+1) by A5,JORDAN3:def 3;
  then
A8: L_Cut(f,p) = <*p*>^g by JORDAN3:def 3;
  per cases;
  suppose
    g is empty;
    then L_Cut(f,p) = <*p*> by A8,FINSEQ_1:34;
    then len L_Cut(f,p) = 1 by FINSEQ_1:39;
    hence contradiction by A5,TOPREAL1:22;
  end;
  suppose
    g is non empty;
    then L~L_Cut(f,p) = LSeg(p,g/.1) \/ L~g by A8,SPPOL_2:20;
    then
A9: f.1 in LSeg(p,g/.1) by A5,A7,XBOOLE_0:def 3;
A10: 1+1 <= len f by A1;
    then
A11: 2 in dom f by FINSEQ_3:25;
    consider i being Nat such that
A12: 1 <= i and
A13: i+1 <= len(<*p*>^g) and
A14: f/.1 in LSeg(<*p*>^g,i) by A5,A3,A8,SPPOL_2:13;
    LSeg(<*p*>^g,i) c= LSeg(f,Index(p,f)+i-'1) by A4,A12,A13,JORDAN3:16;
    then
A15: Index(p,f)+i-'1 = 1 by A1,A14,JORDAN5B:30;
A16: 1 <= Index(p,f) by A4,JORDAN3:8;
    then 1+1 <= Index(p,f)+i by A12,XREAL_1:7;
    then
A17: Index(p,f)+i = 1+1 by A15,XREAL_1:235,XXREAL_0:2;
    then Index(p,f) = 1 by A12,A16,Th6;
    then p in LSeg(f,1) by A4,JORDAN3:9;
    then
A18: p in LSeg(f/.1,f/.(1+1)) by A10,TOPREAL1:def 3;
    i = 1 by A12,A16,A17,Th6;
    then g/.1 = f/.(1+1) by A2,A17,A11,SPRECT_2:8;
    hence contradiction by A6,A3,A9,A18,SPRECT_3:6;
  end;
end;
