reserve X for BCI-algebra;
reserve n for Nat;
reserve x,y for Element of X;
reserve a,b for Element of AtomSet(X);
reserve m,n for Nat;
reserve i,j for Integer;

theorem Th15:
  (a\b)|^n = a|^n\(b|^n)
proof
  defpred P[Nat] means (a\b)|^$1 = a|^$1\(b|^$1);
A1: now
    let n;
A2: (b|^n)` in AtomSet(X) by BCIALG_1:34;
A3: a|^(n+1) in AtomSet(X) by Th13;
    assume P[n];
    then (a\b)|^(n+1) = (a\b)\(a|^n\(b|^n))` by Th2
      .= (a\(a|^n\(b|^n))`)\b by BCIALG_1:7
      .= (a\((a|^n)`\(b|^n)`))\b by BCIALG_1:9
      .= ((b|^n)`\((a|^n)`\a))\b by A2,Th1
      .= ((b|^n)`\b)\((a|^n)`\a) by BCIALG_1:7
      .= (b|^(n+1))`\((a|^n)`\a) by Th14
      .= (b|^(n+1))`\(a|^(n+1))` by Th14
      .= ((b|^(n+1))\(a|^(n+1)))` by BCIALG_1:9
      .= a|^(n+1)\(b|^(n+1)) by A3,BCIALG_1:30;
    hence P[n+1];
  end;
  (a\b)|^0=0.X by Def1
    .=(0.X)` by BCIALG_1:def 5
    .=a|^0\0.X by Def1;
  then
A4: P[0] by Def1;
  for n holds P[n] from NAT_1:sch 2(A4,A1);
  hence thesis;
end;
