reserve N,M,K for ExtNat;

theorem Th10:
  N <= M + 1 implies N <= M or N = M + 1
proof
  assume A1: N <= M + 1;
  per cases by Th3;
  suppose M is Nat;
    then reconsider m = M as Nat;
    m + (1 qua ExtNat) = M + 1;
    then A2: m + 1 in NAT & N <= m + 1 by A1;
    then N in NAT by Th5;
    then reconsider n = N as Nat;
    n <= m or n = m + 1 by A2, NAT_1:8;
    then N <= M or N = m + (1 qua ExtNat);
    hence thesis;
  end;
  suppose M = +infty;
    hence thesis by A1, XXREAL_3:def 2;
  end;
end;
