Cancellation of inequalities in floating point arithmetic


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: Are the inequalities $ x+y<z$ and $ x<z-y$ equivalent in finite precision floating point arithmetic?