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

theorem
  for X being BCI-algebra holds (X is BCI-commutative iff for a being
Element of AtomSet(X),x,y being Element of BranchV(a) holds x\(x\y) = y\(y\x) )
proof
  let X be BCI-algebra;
  thus X is BCI-commutative implies for a being Element of AtomSet(X), x,y
  being Element of BranchV(a) holds x\(x\y) = y\(y\x)
  proof
    assume
A1: X is BCI-commutative;
    let a be Element of AtomSet(X);
    let x,y be Element of BranchV(a);
    y\x in BranchV(a\a) by BCIALG_1:39;
    then ex z1 being Element of X st y\x=z1 & (a\a) <= z1;
    then 0.X <= y\x by BCIALG_1:def 5;
    then 0.X\(y\x) = 0.X;
    then
A2: (y\(y\x))\(x\(x\y)) = 0.X by A1,Th17;
    x\y in BranchV(a\a) by BCIALG_1:39;
    then ex z being Element of X st x\y=z & (a\a) <= z;
    then 0.X <= x\y by BCIALG_1:def 5;
    then 0.X\(x\y) = 0.X;
    then (x\(x\y))\(y\(y\x)) = 0.X by A1,Th17;
    hence thesis by A2,BCIALG_1:def 7;
  end;
  assume
A3: for a being Element of AtomSet(X), x,y being Element of BranchV(a)
  holds x\(x\y) = y\(y\x);
  for x,y being Element of X holds (x\y=0.X implies x = y\(y\x))
  proof
    let x,y be Element of X;
    set aa = 0.X\(0.X\x);
    aa``= aa by BCIALG_1:8;
    then
A4: aa in AtomSet(X) by BCIALG_1:29;
A5: aa\x = (0.X\x)\(0.X\x) by BCIALG_1:7
      .= 0.X by BCIALG_1:def 5;
    then
A6: aa<=x;
    assume
A7: x\y=0.X;
    then aa\y = 0.X by A5,BCIALG_1:3;
    then aa<=y;
    then consider aa be Element of AtomSet(X) such that
A8: aa<=x & aa<= y by A4,A6;
    x in BranchV(aa) & y in BranchV(aa) by A8;
    then x\(x\y) = y\(y\x) by A3;
    hence thesis by A7,BCIALG_1:2;
  end;
  hence thesis;
end;
