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 associative BCI-algebra iff
  for x,y,z being Element of X holds (x\y)\(x\z)=y\z & 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
  (x\y)\(x\z)=y\z & x\0.X=x
  proof
    assume
A1: X is associative BCI-algebra;
    let x,y,z be Element of X;
    (y\x)\(z\x)=z\y by A1,Th49;
    then (x\y)\(z\x)=z\y by A1,Th48;
    then (x\y)\(x\z)=z\y by A1,Th48;
    hence thesis by A1,Th48,Th49;
  end;
  assume
A2: for x,y,z being Element of X holds (x\y)\(x\z)=y\z & x\0.X=x;
  for x,y,z being Element of X holds (x\y)\(x\z)=z\y & x`=x
  proof
    let x,y,z be Element of X;
A3: for x,y being Element of X holds y\x=x\y
    proof
      let x,y be Element of X;
      (x\0.X)\(x\0.X)=(0.X)` by A2;
      then x\(x\0.X)=(0.X)` by A2;
      then x\x=(0.X)` by A2;
      then
A4:   x\x=0.X by A2;
      (x\y)\(x\x)=y\x by A2;
      hence thesis by A2,A4;
    end;
    (x\y)\(x\z)=y\z & x\0.X=x by A2;
    hence thesis by A3;
  end;
  hence thesis by Th50;
end;
