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

theorem Th6:
  (A1 (\) A2) ^\k = (A1 ^\k) (\) (A2 ^\k)
proof
  let n be Element of NAT;
  thus ((A1 (\) A2) ^\k).n = (A1 (\) A2).(n+k) by NAT_1:def 3
    .= A1.(n+k) \ A2.(n+k) by Def3
    .= (A1 ^\k).n \ A2.(n+k) by NAT_1:def 3
    .= (A1 ^\k).n \ (A2 ^\k).n by NAT_1:def 3
    .= ((A1 ^\k) (\) (A2 ^\k)).n by Def3;
end;
