reserve G for strict Group,
  a,b,x,y,z for Element of G,
  H,K for strict Subgroup of G,
  p for Element of NAT,
  A for Subset of G;

theorem Th13:
  b is being_of_order_0 iff for n being Integer st b |^n = 1_G holds n = 0
proof
  thus b is being_of_order_0 implies for n being Integer st b |^n = 1_G
  holds n = 0
  proof
    assume
A1: b is being_of_order_0;
A2: for m being Nat holds b |^ - m = 1_G implies m = 0
    proof
      let m be Nat;
      assume
A3:   b |^ - m = 1_G;
      b |^ -m = (b |^ m)" by GROUP_1:36;
      then
A4:   b" |^ m = 1_G by A3,GROUP_1:37;
      b" is being_of_order_0 by A1,Th12;
      hence thesis by A4,GROUP_1:def 10;
    end;
    for n being Integer holds b |^ n = 1_G implies n = 0
    proof
      let n be Integer;
      assume
A5:   b |^ n = 1_G;
      consider k being Nat such that
A6:   n = k or n = - k by INT_1:2;
      per cases by A6;
      suppose n = k;
        hence thesis by A1,A5,GROUP_1:def 10;
      end;
      suppose
A7:     n = -k;
        then k = 0 by A2,A5;
        hence thesis by A7;
      end;
    end;
    hence thesis;
  end;
  assume for n being Integer holds b|^n = 1_G implies n = 0;
  then for n being Nat holds b |^n = 1_G implies n = 0;
  hence thesis by GROUP_1:def 10;
end;
