reserve m,n for Nat;
reserve i,j for Integer;
reserve S for non empty addMagma;
reserve r,r1,r2,s,s1,s2,t,t1,t2 for Element of S;
reserve G for addGroup-like non empty addMagma;
reserve e,h for Element of G;
reserve G for addGroup;
reserve f,g,h for Element of G;
reserve u for UnOp of G;
reserve A for Abelian addGroup;
reserve a,b for Element of A;
reserve x for object;
reserve y,y1,y2,Y,Z for set;
reserve k for Nat;
reserve G for addGroup;
reserve a,g,h for Element of G;
reserve A for Subset of G;
reserve G for non empty addMagma,
  A,B,C for Subset of G;
reserve a,b,g,g1,g2,h,h1,h2 for Element of G;
reserve G for addGroup-like non empty addMagma;
reserve h,g,g1,g2 for Element of G;
reserve A for Subset of G;
reserve H for Subgroup of G;
reserve h,h1,h2 for Element of H;
reserve G,G1,G2,G3 for addGroup;
reserve a,a1,a2,b,b1,b2,g,g1,g2 for Element of G;
reserve A,B for Subset of G;
reserve H,H1,H2,H3 for Subgroup of G;
reserve h,h1,h2 for Element of H;
reserve x,y,y1,y2 for set;
reserve G for addGroup;
reserve a,b,c,d,g,h for Element of G;
reserve A,B,C,D for Subset of G;
reserve H,H1,H2,H3 for Subgroup of G;
reserve n for Nat;
reserve i for Integer;
reserve L for Subset of Subgroups G;
reserve N2 for normal Subgroup of G;

theorem
  for H being Subgroup of G holds H is normal Subgroup of G iff for A
  holds A + H = H + A
proof
  let H be Subgroup of G;
  thus H is normal Subgroup of G implies for A holds A + H = H + A
  proof
    assume
A1: H is normal Subgroup of G;
    let A;
    thus A + H c= H + A
    proof
      let x be object;
      assume x in A + H;
      then consider a,h such that
A2:   x = a + h and
A3:   a in A and
A4:   h in H by Th94;
      x in a + H by A2,A4,Th103;
      then x in H + a by A1,Th117;
      then ex g st x = g + a & g in H by Th104;
      hence thesis by A3;
    end;
    let x be object;
    assume x in H + A;
    then consider h,a such that
A5: x = h + a & h in H and
A6: a in A by ThB95;
    x in H + a by A5,Th104;
    then x in a + H by A1,Th117;
    then ex g st x = a + g & g in H by Th103;
    hence thesis by A6;
  end;
  assume
A7: for A holds A + H = H + A;
  now
    let a;
    thus a + H = {a} + H .= H + {a} by A7
      .= H + a;
  end;
  hence thesis by Th117;
end;
