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

theorem Th43:
  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)\((a\x)\x) = 0.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)\((a\x
  )\x) = 0.X
  proof
    assume
A2: X is BCK-implicative;
    let x be Element of X;
    (a\x)\((a\x)\x) = x\(x\(a\x)) by Def1
      .= x\x by A2;
    hence thesis by BCIALG_1:def 5;
  end;
  assume
A3: for x being Element of X holds (a\x)\((a\x)\x) = 0.X;
  for x,y being Element of X holds x\(y\x)=x
  proof
    let x,y be Element of X;
    (a\x)\((a\x)\x) = 0.X by A3;
    then
A4: 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
A5: 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 A4,BCIALG_1:def 7;
    then
A6: x\(x\(y\x)) = 0.X by A5;
    (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 A6,BCIALG_1:def 7;
  end;
  hence thesis;
end;
