reserve i,j,n,n1,n2,m,k,u for Nat,
        r,r1,r2 for Real,
        x,y for Integer,
        a,b for non trivial Nat;

theorem Th35:
  n1 <= 4*n & n2 <= 4*n & |.x.| = Py(a,n1) & |.y.| = Py(a,n2) &
    x,y are_congruent_mod Px(a,n)
  implies n1,n2 are_congruent_mod 2*n or n1,-n2 are_congruent_mod 2*n
proof
  assume A1: n1 <= 4*n & n2 <= 4*n & |.x.| = Py(a,n1) & |.y.| = Py(a,n2) &
         x,y are_congruent_mod Px(a,n);
  then
A2: y,x are_congruent_mod Px(a,n) by INT_1:14;
  per cases;
  suppose n1 <= 2*n & n2 <= 2*n or n1 > 2*n & n2 > 2*n or n1 <= 2*n < n2;
    hence thesis by A1,Th34,Lm10,Lm11;
  end;
  suppose  n2 <=2*n <n1;
    then n2,n1 are_congruent_mod 2*n or n2,-n1 are_congruent_mod 2*n
    by A1,A2,Lm10;
    then n1,n2 are_congruent_mod 2*n or (-1)*n2,(-1)*(-n1)
    are_congruent_mod 2*n by INT_1:14,INT_4:11;
    hence thesis by INT_1:14;
  end;
end;
