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 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\(x\y))
  \(y\x)=y\(y\x) )
proof
  let X be non empty BCIStr_0;
  thus X is 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\(x\y))\(y\x)=y\(y\x) by Def24,Th1,Th2;
  thus (for x,y,z being Element of X holds ((x\y)\(x\z))\(z\y)=0.X & x\0.X=x &
  (x\(x\y))\(y\x)=y\(y\x)) implies X is implicative BCI-algebra
  proof
    assume
A1: for x,y,z being Element of X holds ((x\y)\(x\z))\(z\y)=0.X & x\0.X
    =x &(x\(x\y))\(y\x)=y\(y\x);
A2: now
      let x,y,z be Element of X;
      thus ((x\y)\(x\z))\(z\y)=0.X by A1;
      (x\(x\y))\y=((x\0.X)\(x\y))\y by A1
        .=((x\0.X)\(x\y))\(y\0.X) by A1;
      hence (x\(x\y))\y = 0.X by A1;
    end;
    now
      let x,y be Element of X;
      assume that
A3:   x\y = 0.X and
A4:   y\x = 0.X;
      x=x\0.X by A1
        .=(y\(y\x))\(x\y) by A1,A3
        .=y\0.X by A1,A3,A4;
      hence x=y by A1;
    end;
    then
A5: X is being_BCI-4;
    now
      let x be Element of X;
      x\x=(x\0.X)\x by A1
        .=(x\0.X)\(x\0.X) by A1
        .=((x\0.X)\(x\0.X))\0.X by A1
        .=((x\0.X)\(x\0.X))\(0.X)` by A1;
      hence x\x=0.X by A1;
    end;
    then X is being_I;
    hence thesis by A1,A5,A2,Def24,Th1;
  end;
end;
