Floating Point – Proof that a guard digit limits the subtraction error

I read what all computer scientists should know about floating point arithmetic, which is extremely interesting. But I have some difficulty understanding the proof of Theorem 9 (page 33).

First of all a fairly trivial question. When the formula $ (15) $ say:
$$ y – bar {y} lt ( beta – 1) ( beta ^ {- p} + points + beta ^ {- p-k}) $$
It should not be $ the $ instead of $ lt $, or did I miss something?

Most importantly, I do not understand why it is said that if $ x- bar {and} lt $ 1then $ delta = 0 $. How can there be no rounding error?

It is said that:
$$ x – y ge 1.0 – 0. overbrace {0 dots 0} ^ k overbrace { rho dots rho} ^ k ; textrm {with} ; rho = beta – 1 $$

Why is it so? Can not the difference be arbitrarily small or even $ 0?

And all in all, correct me if I'm wrong, but the guard figure saves the day only if the entry floats $ x $ and $ y $ are the exact numbers we want to subtract. If they are the exact result of another calculation, we could still have a catastrophic cancellation. The guard figure would only put the calculated value in the correct range.