reserve a,b,c,d,x,j,k,l,m,n for Nat,
  p,q,t,z,u,v for Integer,
  a1,b1,c1,d1 for Complex;

theorem
  c|^m >= a|^m + b|^m implies
  c|^(k+m) >= a|^(k+m) + b|^(k+m)
  proof
    assume
    A0: c|^m >= a|^m + b|^m; then
    A0a: m <> 0 by Lm0;
    A0c: k = 0 implies thesis by A0;
    per cases;
    suppose
      a = 0 or b = 0; then
      A1: (a|^m = 0 or b|^m = 0) & (a|^(k+m) = 0 or b|^(k+m) = 0)
        by A0a,NAT_1:14,NEWTON:11; then
      c >= a & c >= b by A0,A0a,NAT_1:14,PREPOWER:10;
      hence thesis by A1,Lm3a;
    end;
    suppose
      a > 0 & b > 0;
      hence thesis by A0,A0c,Th18;
    end;
  end;
