 reserve i,j, k,v, w for Nat;
 reserve j1,j2, m, n, s, t, x, y for Integer;
 reserve p for odd Prime;

theorem lem2:
  LAG4SQg v is one-to-one
    proof
      for n1, n2 be object st n1 in dom LAG4SQg v & n2 in dom LAG4SQg v
      & (LAG4SQg v).n1 = (LAG4SQg v).n2 holds n1 = n2
      proof
        let n1, n2 be object such that
A1:     n1 in dom LAG4SQg v and
A2:     n2 in dom LAG4SQg v and
A3:     (LAG4SQg v).n1 = (LAG4SQg v).n2;
A4:     dom LAG4SQg v = Seg len LAG4SQg v by FINSEQ_1:def 3
                       .= Seg v by Def3;
        consider m1 be Nat such that
A5:     n1 = m1 and
A6:     1 <= m1 and
        m1 <= v by A1,A4;
        consider m2 be Nat such that
A7:     n2 = m2 and
A8:     1 <= m2 and
        m2 <= v by A2,A4;
        (LAG4SQg v).m1 = (LAG4SQg v).m2 implies m1 = m2
        proof
          assume
A11:      (LAG4SQg v).m1 = (LAG4SQg v).m2;
          assume
A12:      m1 <> m2;
A13:      (LAG4SQg v).m1 = -1- (m1 - 1)^2 by Def3,A5,A1;
          (LAG4SQg v).m2 =-1 - (m2 - 1)^2 by Def3,A2,A7; then
A14:      (LAG4SQg v).m1 - (LAG4SQg v).m2 = (m2 + m1 - 2)*(m2 -m1) by A13;
A16:      m2 + m1 -2 > 0
          proof
            per cases by A8,XXREAL_0:1;
              suppose m2 = 1; then
A18:            m1 > 1 by A6,A12,XXREAL_0:1;
A19:            m1 + m2 > 1 + 1 by A8,A18, XREAL_1:8;
                m1 + m2 + (-2) > 2 + (-2) by A19,XREAL_1:8;
                hence thesis;
              end;
              suppose m2 >1; then
A20:            m1 + m2 > 1 + 1 by A6,XREAL_1:8;
                m1 + m2 + (-2) > 2 + (-2) by A20,XREAL_1:8;
                hence thesis;
              end;
          end;
          m2 - m1 <> 0 by A12;
          hence contradiction by A11,A14,A16;
          end;
          hence thesis by A5,A7,A3;
        end;
        hence thesis by FUNCT_1:def 4;
      end;
