Assume the valuation of x is strictly greater than the valuation of y. This means 1+x/y is in R, but not 1+y/x. Hence y ≤ x+y, but not so for x.
Since R does not contain 1+y/x it contains the inverse, x/(x+y). It also contains 1, hence R contains y/(x+y). The sum is both ≤ and ≥ y, hence it equals y. If x and y have different valuations their sum has the lesser valuation.
Recall the example of the last section, fractions without p in the denomiinator. The valuation is the exponent of p. Sure enough, p2 + p4 has exactly two powers of p in its factorization. The valuation of the sum is 2. When the operands have the same valuation the sum could have the same valuation or larger. Add p+p to get the same valuation, or p + (p-1)p to get a larger valuation.