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

theorem Th46:
  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 being Element of
  X holds (a\y)\((a\y)\x) = x\y )
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 being Element of X holds (a\y)\((a
  \y)\x) = x\y
  proof
    assume
A2: X is BCK-implicative;
    let x,y be Element of X;
    X is involutory by A2,Th37;
    then
A3: x\(a\y) = y\(a\x) by A1,Th23;
A4: (a\y)\((a\y)\x) = x\(x\(a\y)) by Def1;
    (y\(a\x))\y = (y\y)\(a\x) by BCIALG_1:7
      .= ((a\x))` by BCIALG_1:def 5
      .= 0.X by BCIALG_1:def 8;
    then x\(a\y) <= y by A3;
    then x\y <= (a\y)\((a\y)\x) by A4,BCIALG_1:5;
    then
A5: (x\y)\((a\y)\((a\y)\x)) = 0.X;
    X is BCK-positive-implicative BCK-algebra by A2,Th34;
    then (a\y)= (a\y)\y by Th28;
    then ((a\y)\((a\y)\x))\(x\y) = 0.X by BCIALG_1:1;
    hence thesis by A5,BCIALG_1:def 7;
  end;
  assume
A6: for x,y being Element of X holds (a\y)\((a\y)\x) = x\y;
  for x being Element of X holds (a\x)\((a\x)\x) = 0.X
  proof
    let x be Element of X;
    (a\x)\((a\x)\x) = x\x by A6;
    hence thesis by BCIALG_1:def 5;
  end;
  hence thesis by A1,Th43;
end;
