reserve a,b for object, I,J for set;
reserve b for bag of I;
reserve R for asymmetric transitive non empty RelStr,
  a,b,c for bag of the carrier of R,
  x,y,z for Element of R;
reserve p for partition of b-'a, q for partition of b;
reserve J for set, m for bag of I;

theorem
  for R being asymmetric Relation of I
  for p,q,r being I-valued FinSequence holds
  [p,q] in LexOrder(I,R) iff [r^p,r^q] in LexOrder(I,R)
  proof
    let R be asymmetric Relation of I;
    let p,q,r be I-valued FinSequence;
    hereby
      assume [p,q] in LexOrder(I,R);
      then per cases by LO;
      suppose p c< q;
        then r^p c= r^q & r^p <> r^q by FINSEQ_6:13,FINSEQ_1:33,XBOOLE_0:def 8;
        then r^p c< r^q by XBOOLE_0:def 8;
        hence [r^p,r^q] in LexOrder(I,R) by LO;
      end;
      suppose
        ex k being Nat st k in dom p & k in dom q & [p.k,q.k] in R &
        for n being Nat st 1 <= n < k holds p.n = q.n;
        then consider k being Nat such that
A1:     k in dom p & k in dom q & [p.k,q.k] in R &
        for n being Nat st 1 <= n < k holds p.n = q.n;
        set i = len r; set j = i+k;
A2:     j in dom(r^p) & j in dom(r^q) by A1,FINSEQ_1:28;
A3:     (r^p).j = p.k & (r^q).j = q.k by A1,FINSEQ_1:def 7;
        for n being Nat st 1 <= n < j holds (r^p).n = (r^q).n
        proof let n be Nat;
          assume
A4:       1 <= n < j;
          per cases  by NAT_1:13;
          suppose n <= i;
            then n in dom r by A4,FINSEQ_3:25;
            then (r^p).n = r.n = (r^q).n by FINSEQ_1:def 7;
            hence thesis;
          end;
          suppose i+1 <= n;
            then consider m being Nat such that
A5:         n = i+1+m by NAT_1:10;
A5A:        n = i+(1+m) by A5;
            then
A6:         1 <= 1+m < k by A4,NAT_1:11,XREAL_1:6;
            k <= len p & k <= len q by A1,FINSEQ_3:25;
            then 1 <= 1+m <= len p & 1+m <= len q by A6,XXREAL_0:2;
            then 1+m in dom p & 1+m in dom q by FINSEQ_3:25;
            then (r^p).n = p.(1+m) & (r^q).n = q.(1+m) by A5A,FINSEQ_1:def 7;
            hence thesis by A1,A6;
          end;
        end;
        hence [r^p,r^q] in LexOrder(I,R) by LO,A1,A2,A3;
      end;
    end;
    assume [r^p,r^q] in LexOrder(I,R);
    then per cases by LO;
    suppose r^p c< r^q;
      then p c< q by Lem13;
      hence [p,q] in LexOrder(I,R) by LO;
    end;
    suppose
      ex k being Nat st k in dom(r^p) & k in dom(r^q) &
      [(r^p).k,(r^q).k] in R &
      for n being Nat st 1 <= n < k holds (r^p).n = (r^q).n;
      then consider k being Nat such that
A1:   k in dom(r^p) & k in dom(r^q) & [(r^p).k,(r^q).k] in R &
      for n being Nat st 1 <= n < k holds (r^p).n = (r^q).n;
      set i = len r;
A3:   1 <= k by A1,FINSEQ_3:25;
      now assume k <= i;
        then k in dom r by A3,FINSEQ_3:25;
        then (r^p).k = r.k = (r^q).k by FINSEQ_1:def 7;
        hence contradiction by A1,PREFER_1:22;
      end;
      then k >= i+1 by NAT_1:13;
      then consider j being Nat such that
A4:   k = i+1+j by NAT_1:10;
A5:   k = i+(1+j) by A4;
      k <= len(r^p) = i+len p & k <= len(r^q) = i+len q
      by A1,FINSEQ_3:25,FINSEQ_1:22;
      then
A9:   1 <= 1+j <= len p & 1+j <= len q by A5,NAT_1:11,XREAL_1:6;
      then
A6:   1+j in dom p & 1+j in dom q by FINSEQ_3:25;
      then
A7:   (r^p).k = p.(1+j) & (r^q).k = q.(1+j) by A5,FINSEQ_1:def 7;
      for n being Nat st 1 <= n < 1+j holds p.n = q.n
      proof let n be Nat; assume
A8:     1 <= n < 1+j;
        then n < len p & n < len q by A9,XXREAL_0:2;
        then n in dom p & n in dom q by A8,FINSEQ_3:25;
        then p.n = (r^p).(i+n) & q.n = (r^q).(i+n) & 1 <= i+n < k
        by FINSEQ_1:def 7,NAT_1:12,XREAL_1:6,A5,A8;
        hence thesis by A1;
      end;
      hence thesis by LO,A1,A6,A7;
    end;
  end;
