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 Th24:
  a|^(i+j) = a|^i\(a|^j)`
proof
  per cases;
  suppose
    j>0;
    then reconsider j as Element of NAT by INT_1:3;
    a|^i\(a|^j)`=a|^i\(a`|^j) by Th17
      .=a|^i\(a|^(-j)) by Th10
      .=a|^(i-(-j)) by Th22;
    hence thesis;
  end;
  suppose
A1: j=0;
    then a|^(i+j) = a|^i\0.X by BCIALG_1:2
      .=a|^i\(0.X)` by BCIALG_1:2
      .=a|^i\(a|^j)` by A1,Th3;
    hence thesis;
  end;
  suppose
    j<0;
    then reconsider n=-j as Element of NAT by INT_1:3;
    reconsider b=a` as Element of AtomSet(X) by BCIALG_1:34;
    a|^i\(a|^j)`=a|^i\(a|^(--j))` .=a|^i\(a`|^n)` by Th10
      .=a|^i\(b`|^n) by Th17
      .=a|^i\(a|^n) by BCIALG_1:29
      .=a|^(i-n) by Th22;
    hence thesis;
  end;
end;
