
theorem Th10:
for n,m,b being Nat st n<>0 & b > 1 & m < len digits(n,b) holds n >= b|^m
proof
  let n,m,b be Nat;
  assume A1: n<>0 & b > 1 & m < len digits(n,b);
  A2: value(digits(n,b),b)=n by A1,NUMERAL1:def 2;
  len digits(n,b) > 0 by A1;
  then A3: len digits(n,b) >= 1 by NAT_1:14;
  digits(n,b).((len digits(n,b))-1) <> 0 by A1,NUMERAL1:def 2;
  then A4: digits(n,b).((len digits(n,b))-1) >= 1 by NAT_1:14;
  m+1 <= len digits(n,b) by A1,NAT_1:13;
  then m+1-1 <= (len digits(n,b))-1 by XREAL_1:9;
  then (len digits(n,b))-'1 >= m by A3,XREAL_1:233;
  then A5: b|^((len digits(n,b))-'1) >= b|^m by A1,NAT_6:1;
  for i being Nat st i in dom digits(n,b) holds digits(n,b).i<b
  by A1,NUMERAL1:def 2;
  then A6: (digits(n,b).((len digits(n,b))-1))*(b|^((len digits(n,b))-'1))
  <= n by A1,A2,Th8;
  (b|^((len digits(n,b))-'1)) <=
  (digits(n,b).((len digits(n,b))-1))*(b|^((len digits(n,b))-'1))
  by A4,XREAL_1:151;
  then n >= b|^((len digits(n,b))-'1) by A6,XXREAL_0:2;
  hence n >= b|^m by A5,XXREAL_0:2;
end;
