 reserve R for Ring;
 reserve x, y, y1 for set;
 reserve a, b for Element of R;
 reserve V for LeftMod of R;
 reserve v, w for Vector of V;
 reserve u,v,w for Vector of V;
 reserve F,G,H,I for FinSequence of V;
 reserve j,k,n for Nat;
 reserve f,f9,g for sequence of V;
 reserve R for Ring;
 reserve V, X, Y for LeftMod of R;
 reserve u, u1, u2, v, v1, v2 for Vector of V;
 reserve a for Element of R;
 reserve V1, V2, V3 for Subset of V;
 reserve x for set;
 reserve W, W1, W2 for Submodule of V;
 reserve w, w1, w2 for Vector of W;
 reserve D for non empty set;
 reserve d1 for Element of D;
 reserve A for BinOp of D;
 reserve M for Function of [:the carrier of R,D:],D;
reserve B,C for Coset of W;
 reserve V for LeftMod of R;
 reserve W, W1, W2, W3 for Submodule of V;
 reserve u, u1, u2, v, v1, v2 for Vector of V;
 reserve a, a1, a2 for Element of R;
 reserve X, Y, y, y1, y2 for set;
 reserve C for Coset of W;
 reserve C1 for Coset of W1;
 reserve C2 for Coset of W2;
reserve A1,A2,B for Element of Submodules(V);

theorem Th159:
  for R being Abelian right_zeroed add-associative
    right_complementable non empty addLoopStr,
  a being Element of R, i, j being Element of INT.Ring
  holds (Int-mult-left(R)).(i+j,a)
  = (Int-mult-left(R)).(i,a) + (Int-mult-left(R)).(j,a)
  proof
    let R be Abelian right_zeroed add-associative right_complementable
    non empty addLoopStr,
    a be Element of R, i, j be Element of INT.Ring;
    reconsider ii = i, jj = j as Element of INT;
    per cases;
    suppose A1: i in NAT & j in NAT; then
      reconsider i1=i as Element of NAT;
      reconsider j1=j as Element of NAT by A1;
      thus (Int-mult-left(R)).(i+j,a)
       = (Nat-mult-left(R)).(i1+j1,a) by Def23
      .= (i1+j1)*a
      .= i1*a + j1*a by BINOM:15
      .= (Int-mult-left(R)).(i,a) + (Nat-mult-left(R)).(j1,a) by Def23
      .= (Int-mult-left(R)).(i,a) + (Int-mult-left(R)).(j,a) by Def23;
    end;
    suppose i in NAT & not j in NAT;
      hence (Int-mult-left(R)).(i+j,a) = (Int-mult-left(R)).(i,a)
      +(Int-mult-left(R)).(j,a) by Th158;
    end;
    suppose not i in NAT & j in NAT;
      hence (Int-mult-left(R)).(i+j,a)
      =(Int-mult-left(R)).(i,a) + (Int-mult-left(R)).(j,a) by Th158;
    end;
    suppose not i in NAT & not j in NAT; then
      A3: i < 0 & j < 0 by INT_1:3; then
      reconsider i1=-ii as Element of NAT by INT_1:3;
      reconsider j1=-jj as Element of NAT by A3,INT_1:3;
S1:   (Int-mult-left(R)).(i,a) = (Nat-mult-left(R)).(-i,-a) by A3,Def23
         .= i1*(-a);
S2:   (Nat-mult-left(R)).(-j,-a) = j1*(-a);
      thus (Int-mult-left(R)).(i+j,a) = (Nat-mult-left(R)).(-(i+j),-a)
         by Def23,A3
      .= (i1+j1)*(-a)
      .= i1*(-a) + j1*(-a) by BINOM:15
      .= (Int-mult-left(R)).(i,a) + (Int-mult-left(R)).(j,a)
        by A3,Def23,S1,S2;
    end;
  end;
