Any ideas how I can rewire my brain such that $\varphi \leq \psi$ "obviously" means that $\varphi$ implies $\psi$?

Question

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?

Answer 1

If you can't convince that part of your brain with insight, you could try to trick it with language alone, by inserting truth-bearing adjectives into the "less than":

• $\varphi$ is less evident than $\psi$
• $\varphi$ is less general than $\psi$
• $\varphi$ is less accommodating than $\psi$
• $\varphi$ is less broadly applicable than $\psi$
• $\varphi$ is less often true than $\psi$
• $\varphi$ is less likely than $\psi$
• $\varphi$ is less certain than $\psi$
• $\varphi$ is less justifiable than $\psi$
• $\varphi$ is less obvious than $\psi$

This isn't a good standalone solution, because it's blindly fighting language with language. You still need some guiding principle to help you avoid truth-forming adjectives like "strong", "strict", "interesting", "informative", or "useful". Maybe you can use one of the other answers to kick out the misleading adjectives, and then you can cement the victory by replacing it with a better adjective?

Answer 2

Learn how to program. Then, it will become clearer that: $$\rm{false} = 0 < 1 = \rm{true}$$

Answer 3

“$\varphi$ 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 $\varphi\leq\psi$?

Answer 4

Well, weaker statements follow (logically) from stronger statements, just as later elements follow (in order) the earlier ones. In that sense, it's natural that the strongest statement should be first in the tree, while the weakest should be last.

Answer 5

It is the ordering by "amount of evidence for (statement)", or "range of possible situations in which (statement) is true".

If $A \to B$ then any piece of evidence for $A$ is automatically evidence for $B$. This becomes a theorem if the evidence for $A$ is defined as the set of statements that imply $A$, and ordering of evidence is by inclusion of the sets.

If $A \to B$, then any situation where $A$ is true (considering only situations in which the logical rules leading from $A$ to $B$ hold) is one where $B$ is true. If there is a completeness theorem for the logic then having one statement hold in a superset of the models where another holds, is equivalent to syntactic implication of one statement by the other, connecting the idea to the previous definition.

Answer 6

You can think of $\varphi \leq \psi$ as a sort of containment, $\varphi \subset \psi$. That is, $\varphi$ is a special case of $\psi$, so the former implies the latter.