reserve E for RealLinearSpace;
reserve A, B, C for binary-image of E;
reserve a, b, v for Element of E;
reserve F, G for binary-image-family of E;
reserve A, B, C for non empty binary-image of E;

theorem Th8:
  A(-)B = {v where v is Element of E: v + (-1)*B c= A}
  proof
    thus A(-)B c= {v where v is Element of E: v + (-1)*B c= A}
    proof let x be object;
      assume x in A(-)B; then
      consider z be Element of E such that
      A1: x = z & for b be Element of E st b in B holds z - b in A;
      z + (-1)*B c= A
      proof
        let y be object;
        assume y in z + (-1)*B;
        then consider nb be Element of E such that
        A2: y = z + nb & nb in (-1)*B;
        consider b be Element of E such that
        A3: nb = (-1)*b & b in B by A2;
        z - b in A by A3,A1;
        hence y in A by A2,A3,RLVECT_1:16;
      end;
      hence x in {v where v is Element of E: v + (-1)*B c= A} by A1;
    end;
    let x be object;
      assume x in {v where v is Element of E: v + (-1)*B c= A};
      then consider v be Element of E such that
      A4: x = v & v + (-1)*B c= A;
      for b be Element of E st b in B holds v - b in A
      proof
        let b be Element of E;
        assume b in B;
        then (-1)*b in (-1)*B;
        then v + (-1)*b in v + (-1)*B;
        then v-b in v + (-1)*B by RLVECT_1:16;
        hence thesis by A4;
      end;
      hence x in A(-)B by A4;
  end;
