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

theorem Th20:
  (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 Def9
    .= A \+\ (A1 ^\k).n by NAT_1:def 3
    .= (A (\+\) (A1 ^\k)).n by Def9;
end;
