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
  n>1 & b>0 & a>b implies (a|^n-b|^n)*(a+b) > a|^(n+1) - b|^(n+1)
  proof
    assume
    A1: n>1 & b>0 & a>b;
    A2: a|^(n+1)=a*a|^n & b|^(n+1)=b*b|^n & a|^(m+1)=a*a|^m & b|^(m+1)=b*b|^m
    by NEWTON:6;
    consider m such that
    A3: n = 1 + m by A1,NAT_1:10;
    A3a: m >= 1 by A3,A1,NAT_1:13;
    A3b: a|^m > b|^m by A1,A3a,PREPOWER:10;
    A3c: a|^m - b|^m > b|^m - b|^m by A3b, XREAL_1:9;
    (a|^n-b|^n)*(a+b) = a|^(n+1) - b|^(n+1) - a*b|^n + b*a|^n by A2
    .= a|^(n+1) - b|^(n+1) - a*(b|^(m+1)) + b*(a*a|^m) by NEWTON:6,A3
    .= a|^(n+1) - b|^(n+1) - a*(b*b|^m) + b*(a*a|^m) by NEWTON:6
    .= a|^(n+1) - b|^(n+1) + a*b*(a|^m - b|^m);
    hence thesis by A1,A3c,XREAL_1:39;
  end;
