reserve X for BCI-algebra;
reserve x,y,z,u,a,b for Element of X;
reserve IT for non empty Subset of X;

theorem
  for X being non empty BCIStr_0 holds (X is positive-implicative
BCI-algebra iff for x,y,z being Element of X holds ((x\y)\(x\z))\(z\y)=0.X & x\
  0.X=x & x\y=((x\y)\y)\y` & (x\(x\y))\(y\x)=((y\(y\x))\(y\x))\(x\y) )
proof
  let X be non empty BCIStr_0;
  thus X is positive-implicative BCI-algebra implies for x,y,z being Element
of X holds ((x\y)\(x\z))\(z\y)=0.X & x\0.X=x & x\y=((x\y)\y)\y` & (x\(x\y))\(y\
  x)=((y\(y\x))\(y\x))\(x\y) by Lm24,Th1,Th2,Th84;
  assume
A1: for x,y,z being Element of X holds ((x\y)\(x\z))\(z\y)=0.X & x\0.X=x
  & x\y=((x\y)\y)\y` & (x\(x\y))\(y\x)=((y\(y\x))\(y\x))\(x\y);
  now
    let x be Element of X;
    ((x\0.X)\(x\0.X))\(0.X)`=0.X by A1;
    then ((x\0.X)\(x\0.X))\0.X=0.X by A1;
    then ((x\0.X)\x)\0.X=0.X by A1;
    then (x\x)\0.X=0.X by A1;
    hence x\x=0.X by A1;
  end;
  then
A2: X is being_I;
  now
    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)\0.X by A1;
    then (x\0.X)=((y\0.X)\0.X)\0.X by A1;
    then x=((y\0.X)\0.X)\0.X by A1;
    then x=(y\0.X)\0.X by A1;
    then x=y\0.X by A1;
    hence x=y by A1;
  end;
  then
A3: X is being_BCI-4;
  now
    let x,y,z be Element of X;
    thus ((x\y)\(x\z))\(z\y)=0.X by A1;
    ((x\0.X)\(x\y))\(y\0.X)=0.X by A1;
    then (x\(x\y))\(y\0.X)=0.X by A1;
    hence (x\(x\y))\y = 0.X by A1;
  end;
  then X is weakly-positive-implicative BCI-algebra by A1,A2,A3,Th1,Th84;
  hence thesis by Lm25;
end;
