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If I say that $A$ is stronger than $B$, do I mean that $A \Rightarrow B$, or that $B \Rightarrow A$? (Or something else?)

I feel like I have seen both usages in literature, which is confusing.

Thoughts based on intuition:

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$A \Rightarrow B$ means $A$ is a special case of $B$ — $B$ is more general. This would seem to imply that $B$ is “stronger”. (Example: $n$ is an integer implies $n$ is a real number.)

$A \Rightarrow B$ also means that whenever $A$ holds, $B$ must hold. This would seem to imply that $A$ is “stronger”.

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If $A\Rightarrow B$, then for every $C$, if $B\Rightarrow C$ we have that $A\Rightarrow C$.

Therefore $A$ implies *at least* the same propositions that $B$ implies.

We have two options from here:

- $B\Rightarrow A$, in which case $A$ is
**equivalent**to $B$, and they imply the same things. - $B\nRightarrow A)$, that is $B$ does
*not*imply $A$. We have if so that $A$ is*stronger*than $B$ since $A\implies A$, but $B$ does not.

In essence “$A$ is *stronger* than $B$” is when $\{C\mid B\Rightarrow C\}\subsetneq\{C\mid A\Rightarrow C\}$, and *equivalent* is when the sets are equal.

Let’s make the simplifying assumption that $\lnot(B \implies A)$.

Then $A \implies B$ can be informally expressed as “$A$ is (strictly) stronger than $B$.”

It is certainly possible that in this situation, at some time, someone has instead written “$B$ is stronger than $A$.” Stuff happens. We all have written $x<y$ when we meant $y<x$. And interchange of “necessary condition” and “sufficient condition” happens so (relatively) often that it may be best to avoid these terms.

But “$A$ is stronger than $B$” has only one correct interpretation in terms of the direction of the implication (with disagreement, possibly, in the case of equivalence.)

**However**, suppose that we have proved theorem $X$, $\:$(a) under the assumption $A$, and $\:$(b) under the assumption $B$. Then the result (b) is considered to be a **stronger** result than the result (a). That is perfectly consistent with the ordinary meaning of “stronger,” since

$$(B \implies X) \implies (A\implies X).$$

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