reserve n for Nat;

theorem Th1:
  for n,m be Nat holds n <= m & n <> 0 implies (n+1)/n >= (m+1)/m
proof
  let n,m be Nat;
  assume that
A1: n <= m and
A2: n <> 0;
A3: n > 0 by A2;
A4: 1/n >= 1/m by A1,A2,XREAL_1:85;
A5: (n+1)/n = n/n + 1/n .= 1 + 1/n by A2,XCMPLX_1:60;
  (m+1)/m = m/m + 1/m .= 1 + 1/m by A1,A3,XCMPLX_1:60;
  hence thesis by A4,A5,XREAL_1:7;
end;
