reserve k,n,m,l,p for Nat;
reserve n0,m0 for non zero Nat;

theorem Th2:
  n0 is even implies ex k,m st m is odd & k > 0 & n0 = 2|^k * m
proof
  assume
A1: n0 is even;
  set k = 2 |-count n0;
  2|^k divides n0 by NAT_3:def 7;
  then consider m be Nat such that
A2: n0 = 2|^k * m by NAT_D:def 3;
  take k,m;
A3: now
    reconsider m9=m as Element of NAT by ORDINAL1:def 12;
    assume not m is odd;
    then consider m99 be Nat such that
A4: m9 = 2 * m99 by ABIAN:def 2;
    n0 = 2|^k * 2 * m99 by A2,A4
      .= 2|^(k+1)*m99 by NEWTON:6;
    then 2|^(k+1) divides n0 by NAT_D:def 3;
    hence contradiction by NAT_3:def 7;
  end;
  hence m is odd;
  now
    assume k = 0;
    then n0 = 1*m by A2,NEWTON:4;
    hence contradiction by A1,A3;
  end;
  hence k > 0;
  thus n0 = 2|^k * m by A2;
end;
