reserve X for BCK-algebra;
reserve x,y for Element of X;
reserve IT for non empty Subset of X;

theorem Th8:
  X is commutative BCK-algebra implies for x,y,a being Element of X
  st y <= a holds (a\x)\(a\y) = y\x
proof
  assume
A1: X is commutative BCK-algebra;
  let x,y,a be Element of X;
  assume y <= a;
  then
A2: y\a = 0.X;
  (a\x)\(a\y) = (a\(a\y))\x by BCIALG_1:7
    .= (y\0.X)\x by A1,A2,Def1
    .= y\x by BCIALG_1:2;
  hence thesis;
end;
