In finite precision floating point arithmetic the associative property of addition is not satisfied. This is, it is not always the case that $$(a+b)+c=a+(b+c)$$ Even $a=(a+b)-b$ is not always true.
To prove that $x+y<z$ is equivalent to $x<z-y$ with real numbers we can add $-y$ on both sides of $x+y<z$ to get $(x+y)-y<z-y$ and then from this $x=x+(y-y)<z-y$. But I can't repeat the last step for floating point.
Question 1: Are the inequalities $x+y<z$ and $x<z-y$ equivalent in finite precision floating point arithmetic?
Question 2: Are the inequalities $x+y\leq z$ and $x\leq z-y$ equivalent in finite precision floating point arithmetic?
I was too hasty and updated without checking properly. What I really indented to ask is if the breaking of the equivalence could be as drastic as:
Question 3: Could we have $x+y<z$ and $x>z-y$?
Possibly there are already answers below for this. If so, I will check them when I get a chance and select the answer.
Motivation: There is that interview problem that given an ordered array of numbers and a threshold it asks for the number of pairs of numbers of the array which sum is not larger than the threshold. Possible solutions aside, I wanted to understand the particular point of the dangers of switching between testing $x+y<z$ (and also $\leq$ depending on which variant of the problem is presented) and testing $x<z-y$ (or $\leq$).