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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:

$A \Rightarrow B$ 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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    A stronger **theorem** is one which has a weaker hypothesis and/or a stronger conclusion.2011-07-25
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    In your example "n is an integer implies n is a real number", you are thinking of the predicates involved. The integers are a proper subset of the real numbers. That might be the source of your confusion. When logicians talk about A being stronger than B, they're talking about A and B as distinct statements without regard to the predicates they contain, as Asaf explains.2011-07-25
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    Thanks for the explanations. So when I say "A is stronger than B", precisely what can A and B be? Only theorems? Logical sentences?2011-07-25
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    Let me also share the following [Google Buzz that Terence Tao wrote](https://profiles.google.com/114134834346472219368/posts/jTJUyaibC8v) which addresses some of these issues.2011-07-25
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    @bo1024: It means $A$ implies more sentences, then $B$. This is equivalent to $A \Rightarrow B$ because $\Rightarrow$ is a preorder.2011-07-26

2 Answers 2

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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:

  1. $B\Rightarrow A$, in which case $A$ is equivalent to $B$, and they imply the same things.
  2. $B\nRightarrow A)$, 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.

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    I am being somewhat inaccurate for didactic reasons, and I am not mentioning any underlying theory which may be required for some of the implications.2011-07-25
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    Could you add an example or two?2017-12-28
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    Take the language of equality. The empty theory is weaker than the theory stating that there are at least two different elements.2017-12-28
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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

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