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 Th49:
  for X being non empty BCIStr_0 holds (X is associative
BCI-algebra iff for x,y,z being Element of X holds (y\x)\(z\x)=z\y & x\0.X=x )
proof
  let X be non empty BCIStr_0;
  thus X is associative BCI-algebra implies for x,y,z being Element of X holds
  (y\x)\(z\x)=z\y & x\0.X=x
  proof
    assume
A1: X is associative BCI-algebra;
    let x,y,z be Element of X;
    (z\y)\((y\x)\(z\x))=((z\y)\(y\x))\(z\x) by A1,Def20;
    then (z\y)\((y\x)\(z\x)) = ((z\y)\(z\x))\(y\x) by A1,Th7;
    then (z\y)\((y\x)\(z\x)) = ((z\y)\(z\x))\(x\y) by A1,Th48;
    then
A2: (z\y)\((y\x)\(z\x)) = 0.X by A1,Th1;
    ((y\x)\(z\x))\(z\y)=((y\x)\(z\x))\(y\z) by A1,Th48;
    then ((y\x)\(z\x))\(z\y)=0.X by A1,Def3;
    hence thesis by A1,A2,Def7,Th2;
  end;
  thus (for x,y,z being Element of X holds (y\x)\(z\x)=z\y & x\0.X=x) implies
  X is associative BCI-algebra
  proof
    assume
A3: for x,y,z being Element of X holds (y\x)\(z\x)=z\y & x\0.X=x;
A4: for x,y being Element of X holds y\x=x\y
    proof
      let x,y be Element of X;
      (y\0.X)\(x\0.X)=x\y by A3;
      then y\(x\0.X)=x\y by A3;
      hence thesis by A3;
    end;
A5: for a being Element of X holds a\a=0.X
    proof
      let a be Element of X;
      a`\a`=(0.X)` by A3;
      then (a\0.X)\a`=(0.X)` by A4;
      then (a\0.X)\(a\0.X)=(0.X)` by A4;
      then a\a=(0.X)` by A3;
      hence thesis by A3;
    end;
A6: for x,y,z being Element of X holds ((x\y)\(x\z))\(z\y)=0.X & (x\(x\y)
    )\y = 0.X
    proof
      let x,y,z be Element of X;
      ((x\y)\(x\z))\(z\y)=((y\x)\(x\z))\(z\y) by A4
        .=((y\x)\(z\x))\(z\y) by A4
        .=(z\y)\(z\y) by A3;
      hence ((x\y)\(x\z))\(z\y)=0.X by A5;
      x`\(y\x)=y\0.X by A3;
      then (x\0.X)\(y\x)=y\0.X by A4;
      then (x\0.X)\(x\y)=y\0.X by A4;
      then x\(x\y)=y\0.X by A3;
      then (x\(x\y))\y=y\y by A3;
      hence (x\(x\y))\y = 0.X by A5;
    end;
    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 that
A7:   x\y = 0.X and
      y\x = 0.X;
      x`\(y\x)=y\0.X by A3;
      then (x\0.X)\(y\x)=y\0.X by A4;
      then (x\0.X)\(x\y)=y\0.X by A4;
      then x\(x\y)=y\0.X by A3;
      then y=x\(x\y) by A3
        .=x by A3,A7;
      hence thesis;
    end;
    then
A8: X is being_BCI-4;
    X is being_I by A5;
    then reconsider Y=X as BCI-algebra by A6,A8,Th1;
    Y is associative by A4,Th48;
    hence thesis;
  end;
end;
