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
  for X being non empty BCIStr_1 holds (X is commutative
BCK-Algebra_with_Condition(S) iff for x,y,z being Element of X holds x\(0.X\y)
  = x & (x\z)\(x\y) = (y\z)\(y\x) & (x\y)\z = x\(y*z) )
proof
  let X be non empty BCIStr_1;
  thus X is commutative BCK-Algebra_with_Condition(S) implies for x,y,z being
Element of X holds x\(0.X\y) = x & (x\z)\(x\y) = (y\z)\(y\x) & (x\y)\z = x\(y*z
  )
  proof
    assume
A1: X is commutative BCK-Algebra_with_Condition(S);
    let x,y,z be Element of X;
    (x\(x\y))\z = (y\(y\x))\z by A1,Def9;
    then
A2: (x\z)\(x\y) = (y\(y\x))\z by A1,BCIALG_1:7
      .= (y\z)\(y\x) by A1,BCIALG_1:7;
    0.X\y = y` .= 0.X by A1,BCIALG_1:def 8;
    hence thesis by A1,A2,Th11,BCIALG_1:2;
  end;
  thus (for x,y,z being Element of X holds x\(0.X\y) = x & (x\z)\(x\y) = (y\z)
  \(y\x) & (x\y)\z = x\(y*z) ) implies X is commutative
  BCK-Algebra_with_Condition(S)
  proof
    assume
A3: for x,y,z being Element of X holds x\(0.X\y) = x & (x\z)\(x\y) = (
    y\z)\(y\x) & (x\y)\z = x\(y*z);
A4: for x,y being Element of X holds x\0.X = x
    proof
      let x,y be Element of X;
      0.X\(0.X\0.X) = 0.X by A3;
      hence thesis by A3;
    end;
A5: for x,y being Element of X holds (x\y=0.X & y\x=0.X implies x = y)
    proof
      let x,y be Element of X;
      assume x\y=0.X & y\x=0.X;
      then (x\0.X)\0.X = (y\0.X)\0.X by A3;
      then (x\0.X) = (y\0.X)\0.X by A4
        .= (y\0.X) by A4;
      hence x = (y\0.X) by A4
        .= y by A4;
    end;
A6: for x being Element of X holds x\x = 0.X
    proof
      let x be Element of X;
      x = (x\0.X) by A4;
      then x\x = (0.X\0.X)\(0.X\x) by A3
        .= 0.X\(0.X\x) by A4
        .= 0.X by A3;
      hence thesis;
    end;
A7: for x being Element of X holds 0.X\x = 0.X
    proof
      let x be Element of X;
      0.X = (0.X\x)\(0.X\x) by A6
        .= 0.X\x by A3;
      hence thesis;
    end;
A8: for x,y,z being Element of X holds ((x\y)\(x\z))\(z\y)=0.X
    proof
      let x,y,z be Element of X;
      ((x\y)\(x\z))\(z\y) = ((z\y)\(z\x))\(z\y) by A3
        .= ((z\y)\(z\x))\((z\y)\0.X) by A4
        .= (0.X\(z\x))\(0.X\(z\y)) by A3
        .= 0.X\(0.X\(z\y)) by A7
        .= 0.X by A7;
      hence thesis;
    end;
A9: for x,y,z being Element of X st x\y=0.X & y\z=0.X holds x\z=0.X
    proof
      let x,y,z be Element of X;
      assume that
A10:  x\y=0.X and
A11:  y\z=0.X;
      ((x\z)\(x\y))\(y\z)=0.X by A8;
      then (x\z)\(x\y)=0.X by A4,A11;
      hence thesis by A4,A10;
    end;
A12: for x,y,z being Element of X holds ((x\y)\z)\((x\z)\y) = 0.X
    proof
      let x,y,z be Element of X;
      (((x\y)\z)\((x\y)\(x\(x\z))))\((x\(x\z))\z)=0.X by A8;
      then (((x\y)\z)\((x\y)\(x\(x\z))))\((x\(x\z))\(z\0.X))=0.X by A4;
      then
      (((x\y)\z)\((x\y)\(x\(x\z))))\(((x\0.X)\(x\z))\(z\0.X))=0.X by A4;
      then (((x\y)\z)\((x\y)\(x\(x\z))))\0.X=0.X by A8;
      then
A13:  ((x\y)\z)\((x\y)\(x\(x\z)))=0.X by A4;
      ((x\y)\(x\(x\z)))\((x\z)\y) = 0.X by A8;
      hence thesis by A9,A13;
    end;
A14: for x,y being Element of X holds x\(x\y) = y\(y\x)
    proof
      let x,y be Element of X;
      x\(x\y) = (x\(0.X\y))\(x\y) by A3
        .= (y\(0.X\y))\(y\x) by A3
        .= y\(y\x) by A3;
      hence thesis;
    end;
A15: for x,y,z being Element of X st x\y=0.X holds (x\z)\(y\z)=0.X&(z\y)\(
    z\x)=0.X
    proof
      let x,y,z be Element of X;
      assume
A16:  x\y=0.X;
      ((z\y)\(z\x))\(x\y)=0.X & ((x\z)\(x\y))\(y\z)=0.X by A8;
      hence thesis by A4,A16;
    end;
A17: for x,y,z being Element of X holds ((x\y)\(z\y))\(x\z) = 0.X
    proof
      let x,y,z be Element of X;
      ((x\y)\(x\z))\(z\y) = 0.X by A8;
      then ((x\y)\(z\y))\((x\y)\((x\y)\(x\z))) = 0.X by A15;
      then (((x\y)\(z\y))\(x\z))\(((x\y)\((x\y)\(x\z)))\(x\z)) = 0.X by A15;
      then (((x\y)\(z\y))\(x\z))\((((x\y)\0.X)\((x\y)\(x\z)))\(x\z)) = 0.X by
A4;
      then (((x\y)\(z\y))\(x\z))\((((x\y)\0.X)\((x\y)\(x\z)))\((x\z)\0.X)) =
      0.X by A4;
      then (((x\y)\(z\y))\(x\z))\ 0.X = 0.X by A8;
      hence thesis by A4;
    end;
    for x being Element of X holds x`=0.X by A7;
    hence thesis by A3,A6,A5,A14,A12,A17,Def2,Def9,BCIALG_1:def 3,def 4,def 5
,def 7,def 8;
  end;
end;
