reserve X for BCI-algebra;
reserve x,y,z for Element of X;
reserve i,j,k,l,m,n for Nat;
reserve f,g for sequence of the carrier of X;

theorem Th38:
  X is commutative BCK-algebra iff X is BCI-algebra of 0,0,0,0
proof
  thus X is commutative BCK-algebra implies X is BCI-algebra of 0,0,0,0
  proof
    assume
A1: X is commutative BCK-algebra;
    for x,y being Element of X holds Polynom (0,0,x,y) = Polynom (0,0,y,x)
    proof
      let x,y be Element of X;
A2:   x\(x\y) = y\(y\x) by A1,BCIALG_3:def 1;
      ((x,(x\y)) to_power 1,(y\x)) to_power 0 = (x,(x\y)) to_power 1 by
BCIALG_2:1
        .= y\(y\x) by A2,BCIALG_2:2
        .= (y,(y\x)) to_power 1 by BCIALG_2:2
        .= ((y,(y\x)) to_power 1,(x\y)) to_power 0 by BCIALG_2:1;
      hence thesis;
    end;
    hence thesis by Def3;
  end;
  assume
A3: X is BCI-algebra of 0,0,0,0;
  for x,y being Element of X holds x\(x\y) = y\(y\x)
  proof
    let x,y be Element of X;
A4: Polynom (0,0,x,y) = Polynom (0,0,y,x) by A3,Def3;
    x\(x\y) = (x,(x\y)) to_power 1 by BCIALG_2:2
      .= ((y,(y\x)) to_power 1,(x\y)) to_power 0 by A4,BCIALG_2:1
      .= (y,(y\x)) to_power 1 by BCIALG_2:1
      .= y\(y\x) by BCIALG_2:2;
    hence thesis;
  end;
  hence thesis by A3,Th37,BCIALG_3:def 1;
end;
