theorem
  for m, n being Nat holds m < n implies ex p being Element of NAT
  st n = m+p & 1 <= p
proof
  let m, n be Nat;
  assume
A1: m < n;
  then consider p being Nat such that
A2: n=m+p by NAT_1:10;
  reconsider p as Element of NAT by ORDINAL1:def 12;
  take p;
  thus n = m+p by A2;
  assume p < 1;
  then p < 0+1;
  then p = 0 by NAT_1:13;
  hence contradiction by A1,A2;
end;
