theorem Th12:
  b is being_of_order_0 implies b" is being_of_order_0
proof
  assume
A1: b is being_of_order_0;
  for n being Nat st b" |^ n = 1_G holds n = 0
  proof
    let n be Nat;
    assume b" |^ n = 1_G;
    then (b |^n)" = 1_G by GROUP_1:37;
    then (b |^n)" = (1_G)" by GROUP_1:8;
    then b |^n = 1_G by GROUP_1:9;
    hence thesis by A1,GROUP_1:def 10;
  end;
  hence thesis by GROUP_1:def 10;
end;
