
theorem ThDivisible1:
  for V being torsion-free Z_Module holds
  for v being Vector of DivisibleMod(V) holds
  for a being Element of INT.Ring st a <> 0 holds
  ex u being Vector of DivisibleMod(V) st a*u = v
  proof
    let V be torsion-free Z_Module;
    thus for v being Vector of DivisibleMod(V) holds
    for a being Element of INT.Ring st a <> 0 holds
    ex u being Vector of DivisibleMod(V) st a*u = v
    proof
      let v be Vector of DivisibleMod(V);
      assume AS: ex a being Element of INT.Ring st a <> 0 &
      for u being Vector of DivisibleMod(V) holds a*u <> v;
      consider a be Element of INT.Ring such that
      B1: a <> 0 &
      for u being Vector of DivisibleMod(V) holds a*u <> v by AS;
      reconsider vv = v as Element of Class EQRZM(V) by defDivisibleMod;
      set Z = Z_MQ_VectSp(V);
      reconsider Z as VectSp of F_Rat;
      BX: Z = ModuleStr(# Class EQRZM(V), addCoset(V), zeroCoset(V),
      lmultCoset(V) #) by ZMODUL04:def 5;
      reconsider vv as Element of Z by BX;
      reconsider aa = a as Element of F_Rat by NUMBERS:14;
      B2: aa <> 0.F_Rat by B1;
      reconsider ai = aa" as Element of F_Rat;
      B3: aa * ai = 1.F_Rat by B2,VECTSP_1:def 10;
      set uu = ai*vv;
      reconsider uu as Element of Z;
      reconsider u = uu as Element of DivisibleMod(V) by BX,defDivisibleMod;
      aa*uu = (aa*ai)*vv by VECTSP_1:def 16
      .= vv by B3,VECTSP_1:def 17;
      then v = ((lmultCoset(V)) | [:INT, Class EQRZM(V):]).[a,u]
      by BX,FUNCT_1:49,ZFMISC_1:87
      .= a*u by defDivisibleMod;
      hence contradiction by B1;
    end;
  end;
