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,y,z being Element of X
  holds (x\(y\z))\(x\y) <= x\(a\z) )
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,y,z being Element of X holds (x\(y\z
  ))\(x\y) <= x\(a\z)
  proof
    assume
A2: X is BCK-implicative;
    let x,y,z be Element of X;
    X is BCK-positive-implicative BCK-algebra by A2,Th34;
    then
A3: ((x\(y\z))\(x\(x\z)))\((x\y)\z) = ((x\(y\z))\(x\(x\z)))\((x\z)\(y\z))
    by Def11
      .= 0.X by BCIALG_1:1;
    ((x\y)\z)\(x\y) = ((x\y)\(x\y))\z by BCIALG_1:7
      .= z` by BCIALG_1:def 5
      .= 0.X by BCIALG_1:def 8;
    then
A4: ((x\(y\z))\(x\(x\z)))\(x\y) = 0.X by A3,BCIALG_1:3;
    ((x\(y\z))\(x\y))\(x\(a\z)) = ((x\(y\z))\(x\y))\(z\(z\x)) by A1,A2,Th47
      .= ((x\(y\z))\(x\y))\(x\(x\z)) by Def1
      .= 0.X by A4,BCIALG_1:7;
    hence thesis;
  end;
  assume
A5: for x,y,z being Element of X holds (x\(y\z))\(x\y) <= x\(a\z);
  for x being Element of X holds (a\x)\((a\x)\x) = 0.X
  proof
    let x be Element of X;
    (x\(x\x))\(x\x) <= x\(a\x) by A5;
    then (x\(0.X))\(x\x) <= x\(a\x) by BCIALG_1:def 5;
    then x\(x\x) <= x\(a\x) by BCIALG_1:2;
    then x\0.X <= x\(a\x) by BCIALG_1:def 5;
    then x <= x\(a\x) by BCIALG_1:2;
    then x\(x\(a\x)) = 0.X;
    hence thesis by Def1;
  end;
  hence thesis by A1,Th43;
end;
