Any ideas how I can rewire my brain such that $\varphi \leq \psi$ "obviously" means that $\varphi$ implies $\psi$? The Boolean domain $B=\{\mathrm{False},\mathrm{True}\},$ can be viewed as a partially ordered set in two different ways. In the best approach, $\mathrm{False}$ is the least element and $\mathrm{True}$ is the greatest element. This has the effect that the lattice operations $\wedge$ and $\vee$ agree with their usual (logical) definitions; furthermore, we deduce (under this convention) that $\varphi \leq \psi$ iff $\varphi \rightarrow \psi = \mathrm{True},$ as desired. Thus, let us hereafter agree to order the Boolean domain in this way.
The problem is this. Whenever I'm thinking about the Boolean domain, I tend to think of its elements as conditions. So for example, since $\mathrm{True}$ is the weakest condition, I tend to think of $\mathrm{True}$ as the least element. But this contradicts the aforementioned ordering; we agreed that $\mathrm{True}$ should be the greatest element!
The same issue occurs when $B$ is replaced by any other poset of sentences, such as the sentences of the first-order language of arithmetic. I always get confused, because I tend to think of those sentences as conditions, and therefore I read $\varphi \leq \psi$ to mean "$\varphi$ is a weaker condition than $\psi$." (Bad!)
So, I'd like to rewire my brain so that "$\varphi$ is less than $\psi$" intuitively means "$\varphi$ implies $\psi$."
Any ideas how to do this?

“$\varphi$ implies $\psi$” means the following:


*

*If $\varphi$ is true, then $\psi$ must be true. That is, if $\varphi=1$, then $\psi=1$ (using the notation as in the answer of @nbubis).

*If $\varphi$ is false, then $\psi$ may be either false or true. That is, if $\varphi=0$, then either $\psi=0$ or $\psi=1$.


Can you now see why the implication condition is equivalent to requiring that $\varphi\leq\psi$?
