Let $\succeq$ be a preference relation on $X$. Is it true that $x \succeq y$ if and only if $\lnot (y \succ x)$?
I think it is true and my proof is as follows. To prove $\implies$ direction, we have $\lnot (y \succ x) \equiv \lnot(y \succeq x)$ or $x \succeq y$. So clearly if we assume $x \succeq y$, then $\lnot (y \succ x)$ is true. To prove $\impliedby$ direction, assume $\lnot (y \succ x)$ is true. Then either $\lnot(y \succeq x)$ or $x \succeq y$. If $x \succeq y$ then we are done. If $\lnot(y \succeq x)$, then since $\succeq$ is a preference relation it must be complete, which implies $x \succeq y$.
Is my proof correct? Also, in general, I just want to ask, is it true that whenever we have a preference relation $\succeq$, instead of proving all these facts, can we "informally" pretend $\succeq$ is like the inequality $\ge$, and everything works analougously? For example, we could prove that $\lnot (x \sim y)$ if and only if $x \succ y$ or $y \succ x$, but if we just wanted to use deduce this fact without actually proving it, we can clearly see it from $\lnot(x=y)$ if and only if $x>y$ or $y>x$.