theorem
  <%>E in A |^ n & n > 0 implies <%>E in A
proof
  assume that
A1: <%>E in A |^ n and
A2: n > 0;
  consider m such that
A3: m + 1 = n by A2,NAT_1:6;
  A |^ n = (A |^ m) ^^ A by A3,Th23;
  then ex a1, a2 st a1 in A |^ m & a2 in A & a1 ^ a2 = <%>E by A1,Def1;
  hence thesis by AFINSQ_1:30;
end;
