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

theorem
  for X being bounded commutative BCK-algebra,a being Element of X st a
  is being_greatest holds (X is BCK-implicative iff for x being Element of X
  holds (a\x)\x = (a\x) )
proof
  let X be bounded commutative BCK-algebra;
  let a be Element of X;
  assume
A1: a is being_greatest;
  thus X is BCK-implicative implies for x being Element of X holds (a\x)\x = (
  a\x)
  proof
    assume
A2: X is BCK-implicative;
    let x be Element of X;
A3: ((a\x)\x)\(a\x) = ((a\x)\(a\x))\x by BCIALG_1:7
      .= x` by BCIALG_1:def 5
      .= 0.X by BCIALG_1:def 8;
    (a\x)\((a\x)\x) = 0.X by A1,A2,Th43;
    hence thesis by A3,BCIALG_1:def 7;
  end;
  assume
A4: for x being Element of X holds (a\x)\x = (a\x);
  let x,y be Element of X;
  (a\x)\((a\x)\x) = (a\x)\(a\x) by A4
    .= 0.X by BCIALG_1:def 5;
  then
A5: x\(x\(a\x)) = 0.X by Def1;
  y\a = 0.X by A1;
  then y<=a;
  then y\x <= a\x by BCIALG_1:5;
  then
A6: x\(a\x) <= x\(y\x) by BCIALG_1:5;
  (x\(a\x))\x = (x\x)\(a\x) by BCIALG_1:7
    .= (a\x)` by BCIALG_1:def 5
    .= 0.X by BCIALG_1:def 8;
  then x\(a\x) = x by A5,BCIALG_1:def 7;
  then
A7: x\(x\(y\x)) = 0.X by A6;
  (x\(y\x))\x = (x\x)\(y\x) by BCIALG_1:7
    .= (y\x)` by BCIALG_1:def 5
    .= 0.X by BCIALG_1:def 8;
  hence thesis by A7,BCIALG_1:def 7;
end;
