reserve X for non empty BCIStr_1;
reserve d for Element of X;
reserve n,m,k for Nat;
reserve f for sequence of  the carrier of X;

theorem Th46:
  for X being BCK-Algebra_with_Condition(S) holds ( X is
positive-implicative iff for x,y,z being Element of X holds (x*y)\z = (x\z)*(y\
  z) )
proof
  let X be BCK-Algebra_with_Condition(S);
A1: X is positive-implicative implies for x,y,z being Element of X holds (x*
  y)\z = (x\z)*(y\z)
  proof
    assume
A2: X is positive-implicative;
    let x,y,z be Element of X;
    (((x*y)\z)\(x\z))\((x*y)\x) = 0.X by BCIALG_1:def 3;
    then (((x*y)\z)\(x\z)) <= ((x*y)\x);
    then (((x*y)\z)\(x\z))\z <= ((x*y)\x)\z by BCIALG_1:5;
    then
A3: ((((x*y)\z)\(x\z))\z) \ (((x*y)\x)\z) = 0.X;
    (x*y)\x <= y by Lm2;
    then ((x*y)\x)\z <= y\z by BCIALG_1:5;
    then
A4: (((x*y)\x)\z)\(y\z) = 0.X;
    (x*y)\z = (x*y)\(z*z) by A2,Th44
      .= ((x*y)\z)\z by Th11;
    then ((x*y)\z) \ (x\z) = (((x*y)\z)\(x\z))\z by BCIALG_1:7;
    then (((x*y)\z)\(x\z))\(y\z) = 0.X by A3,A4,BCIALG_1:3;
    then
A5: ((x*y)\z)\((x\z)*(y\z)) = 0.X by Th11;
    y <= x*y by Th37;
    then y\z <= (x*y)\z by BCIALG_1:5;
    then ((x*y)\z)*(y\z) <= ((x*y)\z)*((x*y)\z) by Th7;
    then
A6: (((x*y)\z)*(y\z)) \ (((x*y)\z)*((x*y)\z)) = 0.X;
    x <= x*y by Th37;
    then x\z <= (x*y)\z by BCIALG_1:5;
    then (x\z)*(y\z) <= ((x*y)\z)*(y\z) by Th7;
    then ((x\z)*(y\z)) \ (((x*y)\z)*(y\z)) = 0.X;
    then ((x\z)*(y\z)) \ (((x*y)\z)*((x*y)\z)) = 0.X by A6,BCIALG_1:3;
    then ((x\z)*(y\z))\((x*y)\z) = 0.X by A2,Th44;
    hence thesis by A5,BCIALG_1:def 7;
  end;
  (for x,y,z being Element of X holds (x*y)\z = (x\z)*(y\z)) implies X is
  positive-implicative
  proof
    assume
A7: for x,y,z being Element of X holds (x*y)\z = (x\z)*(y\z);
    for x being Element of X holds x*x = x
    proof
      let x be Element of X;
      (x*x)\x = (x\x)*(x\x) by A7;
      then (x*x)\x = 0.X*(x\x) by BCIALG_1:def 5;
      then (x*x)\x = 0.X*0.X by BCIALG_1:def 5;
      then
A8:   (x*x)\x = 0.X by Th8;
      x\(x*x) = (x\x)\x by Th11
        .= x` by BCIALG_1:def 5
        .= 0.X by BCIALG_1:def 8;
      hence thesis by A8,BCIALG_1:def 7;
    end;
    hence thesis by Th44;
  end;
  hence thesis by A1;
end;
