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

theorem Th3:
  n=2|^k & m is odd implies n,m are_coprime
proof
  assume
A1: n=2|^k;
  assume
A2: m is odd;
  then reconsider h = n gcd m as non zero Nat by INT_2:5;
  for p being Element of NAT st p is prime holds p |-count h = p |-count 1
  proof
    reconsider n9=n,m9=m as non zero Nat by A1,A2;
    let p be Element of NAT;
    assume
A3: p is prime;
A4: min(p |-count n,p |-count m) = 0
    proof
      per cases;
      suppose
A5:     p=2;
        not p divides m
        proof
          assume p divides m;
          then consider m9 be Nat such that
A6:       m = p * m9 by NAT_D:def 3;
          thus contradiction by A2,A5,A6;
        end;
        then p |-count m = 0 by A5,NAT_3:27;
        hence thesis by XXREAL_0:def 9;
      end;
      suppose
A7:     p<>2;
        p<>1 by A3,INT_2:def 4;
        then
A8:     p |-count 2 = 0 by A7,INT_2:28,NAT_3:24;
        p |-count n = k * (p |-count 2) by A1,A3,NAT_3:32
          .= k * (0 qua Nat) by A8;
        hence thesis by XXREAL_0:def 9;
      end;
    end;
    p > 1 & p |-count (n9 gcd m9) = min(p |-count n9,p |-count m9) by A3,Lm1,
INT_2:def 4;
    hence p |-count h = p |-count 1 by A4,NAT_3:21;
  end;
  then n gcd m = 1 by NAT_4:21;
  hence n,m are_coprime by INT_2:def 3;
end;
