reserve n,m,k for Nat,
  x,X for set,
  A for Subset of X,
  A1,A2 for SetSequence of X;

theorem Th18:
  (A (\) A1) ^\k = A (\) (A1 ^\k)
proof
  let n be Element of NAT;
  thus ((A (\) A1) ^\k).n = (A (\) A1).(n+k) by NAT_1:def 3
    .= A \ A1.(n+k) by Def7
    .= A \ (A1 ^\k).n by NAT_1:def 3
    .= (A (\) (A1 ^\k)).n by Def7;
end;
