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 Th17:
  c|^m >= a|^m + b|^m & a>0 & b>0 implies
  c|^(m+1) > a|^(m+1) + b|^(m+1)
  proof
    assume
    A0: c|^m >= a|^m + b|^m & a>0 & b>0;
    consider x be Nat such that
    A0b: c = a + x by A0,Th6, NAT_1:10;
    A1: (a+x)|^m >= a|^m + b|^m & a>0 & x>0 by A0,A0b,Lm5d;
    A2a: a|^m > 0 by A0,PREPOWER:6; then
    A3: (a+x)|^m > b|^m by A0,A0b,XREAL_1:39;
    A4:  a+x > b by A3,Lm3a;
    A5: (a+x)|^m*(a+x) >= (a|^m + b|^m)*(a+x) by A0,A0b,XREAL_1:66;
    A7: b|^m > 0 by A0,PREPOWER:6;
    A8: a|^m*(a+x) > a|^m*a by A1,NAT_1:16,A2a,XREAL_1:68;
    b|^m*(a+x) > b|^m*b by A4,A7,XREAL_1:68; then
    a|^m*(a+x) + b|^m*(a+x) > a|^m*a + b|^m*b by A8,XREAL_1:8; then
    A12: (a+x)|^m*(a+x) > a|^m*a + b|^m*b by A5, XXREAL_0:2;
    a|^m*a = a|^(m+1) & b|^m*b = b|^(m+1) by NEWTON:6;
    hence thesis by A0b,A12,NEWTON:6;
  end;
