reserve X for BCK-algebra;
reserve x,y for Element of X;
reserve IT for non empty Subset of X;

theorem
  for X being non empty BCIStr_0 holds (X is BCI-commutative 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\(y\(x\(x\y))) )
proof
  let X be non empty BCIStr_0;
  (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\(y\(x\(x\y)))) implies X is BCI-commutative 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\(y\(x\(x\y)));
    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
A2:   x\y=0.X and
A3:   y\x=0.X;
A4:   x\(x\y) = x by A1,A2;
      y\(y\(x\(x\y))) = y\0.X by A1,A2,A3
        .= y by A1;
      hence thesis by A1,A4;
    end;
    then X is being_BCI-4;
    hence thesis by A1,Th16,BCIALG_1:11;
  end;
  hence thesis by Th16,BCIALG_1:1,2;
end;
