 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 lem1:
  LAG4SQf(v) is one-to-one
  proof
    set f = LAG4SQf v;
    for n1, n2 be object st n1 in dom f & n2 in dom f
    & f.n1 = f.n2 holds n1 = n2
    proof
      let n1, n2 be object such that
A1:   n1 in dom f and
A2:   n2 in dom f and
A3:   f.n1 = f.n2;
A4:   dom f = Seg len f by FINSEQ_1:def 3.= Seg v by Def2;
      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;
      f.m1 = f.m2 implies m1 = m2
      proof
        assume
A11:    f.m1 = f.m2;
        assume
A12:    m1 <> m2;
A13:    f.m1 = (m1 - 1)^2 by Def2,A5,A1;
A14:    f.m1 - f.m2 =
        (m1 - 1)^2 -(m2 - 1)^2 by A13,Def2,A2,A7 .= (m2 + m1 - 2)*(m1 -m2);
A16:    m1 + m2 -2 > 0
        proof
          per cases by A8,XXREAL_0:1;
          suppose m2 = 1; then
            m1 > 1 by A12,A6,XXREAL_0:1; then
A17:        m1 + m2 > 1 + 1 by A8, XREAL_1:8;
            m1 + m2 + (-2) > 2 + (-2) by A17,XREAL_1:8;
            hence m1 + m2 -2 > 0;
          end;
          suppose m2 >1; then
A19:        m1 + m2 > 1 + 1 by A6,XREAL_1:8;
            m1 + m2 + (-2) > 2 + (-2) by A19,XREAL_1:8;
            hence m1 + m2 -2 > 0;
          end;
        end;
        m1 - m2 <> 0 by A12;
        hence contradiction by A11,A16,A14;
      end;
      hence thesis by A3,A5,A7;
    end;
    hence thesis by FUNCT_1:def 4;
  end;
