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

theorem Th19:
  (A1 (\) A) ^\k = (A1 ^\k) (\) A
proof
  let n be Element of NAT;
  thus ((A1 (\) A) ^\k).n = (A1 (\) A).(n+k) by NAT_1:def 3
   .= A1.(n+k) \ A by Def8
   .= (A1 ^\k).n \ A by NAT_1:def 3
   .= ((A1 ^\k) (\) A).n by Def8;
end;
