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

theorem
  X is BCI-weakly-commutative BCI-algebra implies X is BCI-commutative
proof
  assume
A1: X is BCI-weakly-commutative BCI-algebra;
  let x,y be Element of X;
  assume x\y=0.X;
  then (x\0.X)\(0.X\0.X) = y\(y\x) by A1,Def5;
  then (x\0.X)\0.X = y\(y\x) by BCIALG_1:def 5;
  then x\0.X = y\(y\x) by BCIALG_1:2;
  hence thesis by BCIALG_1:2;
end;
