reserve x,y for object,
        i,j,k,m,n for Nat;

theorem Th13:
  for O be odd-valued FinSequence,
      a be natural-valued FinSequence,
      s be odd_organization of O st
        len O = len a & O (#) 2|^a is one-to-one
    holds a*.s.i is one-to-one
proof
  let O be odd-valued FinSequence,
      a be natural-valued FinSequence,
      s be odd_organization of O;
  set p=O (#) (2|^a);
  assume A1:len O = len a & p is one-to-one;
  then reconsider S=s as DoubleReorganization of dom a by FINSEQ_3:29;
  A2:(a*.S).i = a*(S.i) by FLEXARY1:41;
  thus (a*.s).i is one-to-one
  proof
    assume not (a*.s).i is one-to-one;
    then consider x1,x2 be object such that
    A3: x1 in dom ((a*.S).i)& x2 in dom ((a*.S).i)&
    (a*.S).i.x1=(a*.S).i.x2 & x1<>x2;
    reconsider x1,x2 as Nat by A3;
    set s1=s_(i,x1),s2=s_(i,x2);
    A4:x1 in dom (s.i) & s1 in dom a & a.s1 = (a*.S).i.x1
    by A2,A3,FUNCT_1:11,12;
    A5:x2 in dom (S.i) & s2 in dom a & a.s2 = (a*.S).i.x2
    by A2,A3,FUNCT_1:11,12;
    A6:2*i-1 = O.s_(i,1) & ... & 2*i-1 = O.s_(i,len (s.i)) by Def4;
    1<= x1 & x1 <= len (s.i) by A4,FINSEQ_3:25;
    then A7: 2*i-1 = O.s1 by A6;
    1<= x2 & x2 <= len (s.i) by A5,FINSEQ_3:25;
    then A8: 2*i-1 = O.s2 by A6;
    O is (len O)-element by CARD_1:def 7;
    then A9:len p = len a by A1,CARD_1:def 7;
    A10:1<= s.i.x1 & s.i.x1 <= len a by A4,FINSEQ_3:25;
    1<= s.i.x2 & s.i.x2 <= len a by A5,FINSEQ_3:25;
    then A11:s2 in dom p & s1 in dom p by A9,A10,FINSEQ_3:25;
    A12:p.s1 = (O.s1) * ((2|^a).s1) by VALUED_1:5;
    A13:(2|^a).s1 = 2 to_power (a.s1) by A4,FLEXARY1:def 4;
    p.s2 = (O.s2) * ((2|^a).s2) by VALUED_1:5;
    then p.s2 = p.s1 by A5,FLEXARY1:def 4,A13,A12,A7,A8,A4,A3;
    then s2 = s1 by A11,A1;
    hence thesis by A4,A5,FUNCT_1:def 4,A3;
  end;
end;
