reserve ADG for Uniquely_Two_Divisible_Group;
reserve a,b,c,d,a9,b9,c9,p,q for Element of ADG;
reserve x,y for set;

theorem Th13:
  a,b ==> b,c & a,b9 ==> b9,c implies b=b9
proof
  assume a,b ==> b,c & a,b9 ==> b9,c;
  then a + c = b + b & a + c = b9 + b9 by Th5;
  then (b+(-b9))+b = (b9+b9)+(-b9) by RLVECT_1:def 3
    .= b9 +(b9 +(-b9)) by RLVECT_1:def 3
    .= b9 + 0.ADG by RLVECT_1:5
    .= b9 by RLVECT_1:4;
  then
A1: (b+(-b9)) + (b+(-b9)) = b9+ (-b9) by RLVECT_1:def 3
    .= 0.ADG by RLVECT_1:5;
  b9 = 0.ADG + b9 by RLVECT_1:4
    .= (b+(-b9))+b9 by A1,VECTSP_1:def 18
    .= b+((-b9)+b9) by RLVECT_1:def 3
    .= b+0.ADG by RLVECT_1:5
    .= b by RLVECT_1:4;
  hence thesis;
end;
