My professors are always saying, "This theorem is strong" or "There is a way to make a much stronger version of this result" or things like that. In my mind, a strong theorem is able to tell you a lot of important information about something, but this does not seem to be what they mean. What is strength? Is it a formal idea?

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    $\begingroup$ Usually, when I hear that Theorem B is "stronger than Theorem A" in mathematics, I think of that Theorem B uses less stringent assumptions than Theorem A for a similar result. $\endgroup$ Commented Feb 29, 2016 at 17:15
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    $\begingroup$ It's not completely well-defined, but as examples: the less assumptions and the more general the conclusion, the stronger the theorem. Or, another type of "strength": the less assumptions and the tighter (impossible to improve) the conclusion, the stronger the theorems. $\endgroup$
    – Clement C.
    Commented Feb 29, 2016 at 17:16
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    $\begingroup$ Sometimes, it just means that one obviously implies the other, but not visa versa. $\endgroup$ Commented Feb 29, 2016 at 17:22
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    $\begingroup$ It’s a natural transformation $s_{A,B} : A \otimes F(B) \to F(A \otimes B)$, such that… oops, sorry, I guess you didn’t mean the category-theoretic notion. $\endgroup$ Commented Feb 29, 2016 at 17:38
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    $\begingroup$ When a professor is saying this, they mean "You should remember this". $\endgroup$ Commented Mar 1, 2016 at 8:17

6 Answers 6


Suppose you have a theorem that says "If $X$, then $Y$." There are two ways to strengthen such a theorem:

  1. Assume less. If you can reduce the number of hypotheses, but still prove the same conclusion, then you have proved a more "powerful" result (in the sense that it applies in more situations).
  2. Prove more. If you can keep the same hypotheses, but add more information to the conclusion, then you have also produced a more "powerful" result.

Here is an easy example from Geometry.

Let $ABCD$ be a (non-square) rectangle. Then the internal angle bisectors of the vertices intersect at four points $WXYZ$, which are the vertices of a rectangle.

(You need the condition that $ABCD$ is not a square because if it is a square then all four angle bisectors coincide at a single point.)

Here are a few ways to strengthen the theorem:

  1. The hypothesis "$ABCD$ is a (non-square) rectangle" can be relaxed to the more general "$ABCD$ is a (non-rhombic) parallelogram". The conclusion that $WXYZ$ is a rectangle still holds.
  2. Alternatively, you can keep the original hypothesis that $ABCD$ is a (non-square) rectangle, and strengthen to the conclusion to say that $WXYZ$ is not just a rectangle, but a square.
  3. Having done that, you can then strengthen the conclusion of the theorem even more, by noting that the diagonal of square $WXYZ$ is equal in length to the difference of the lengths of the sides of $ABCD$.
  4. Once you know that, you can now strengthen the theorem even more by (finally) removing the hypothesis that $ABCD$ is non-square, and including the case in which the four angle bisectors coincide at a single point as forming a "degenerate" square with a diagonal of length zero.
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    $\begingroup$ This is an excellent answer with a very well-chosen illustration (simple but not too simple). Perhaps it would be nice to see some diagrams but the text is very clear and well-written. $\endgroup$
    – Silverfish
    Commented Feb 29, 2016 at 19:21
  • $\begingroup$ This explains well what a stronger theorem is. But what is a strong theorem? $\endgroup$
    – GOTO 0
    Commented Feb 29, 2016 at 19:39
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    $\begingroup$ @GOTO0 As far as I know the word "strength" is meaningful only in a relative sense. I suppose in general one can say that a theorem is "strong" if it assumes very little and concludes a lot -- but what constitutes "a little" and "a lot" would seem to be very subjective. $\endgroup$
    – mweiss
    Commented Feb 29, 2016 at 19:57
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    $\begingroup$ @mweiss you're right that "strength" is a relative term. If a statement is just called "strong", that should be understood in the context of the whole reasoning. That could mean that the statement is stronger than other statements under discussion, or that it is pretty difficult(impossible) to find or prove a stronger statement. Otherwise, the "strongness" becomes subjective and so less useful for strict reasoning, but may still be useful for educational purposes. $\endgroup$
    – sshilovsky
    Commented Mar 1, 2016 at 8:00
  • $\begingroup$ I sometimes come across statements where "strength" is used strictly of the conclusions; someone will say "we can strengthen this result by using more assumptions". In the sense discussed in this answer, that would be simultaneously weakening and strengthening the theory, which you couldn't call "stronger" without measuring the "amount" of strength traded off in each direction. $\endgroup$
    – Ben
    Commented Mar 2, 2016 at 4:50

Claims are said to be stronger or weaker, depending on the amount of information you can imply from the claim. For example, if $x$ is a positive solution of the equation $x^2 = 2$, then the claim $x > 0$ is weaker than $x > 1$ (the second is more precise).

Sometimes we try to make these implications strongest possible. For example, $1$ is the strongest integer lower bound on such an $x$.


In logic, saying that $\text{statement }x\text{ is stronger than statement }y$, is equivalent to saying that:

$$[\text{statement }x\text{ implies statement }y]\text{ but [statement }y\text{ does not imply statement }x]$$

For example, suppose we conjecture the following statements on a set $S$:

  • Conjecture $x$: all the elements in $S$ are divisible by $4$
  • Conjecture $y$: all the elements in $S$ are divisible by $2$

It is obvious that conjecture $x$ is stronger than conjecture $y$.

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    $\begingroup$ Yes, but in the OP's context, all theorems are de facto true, so this sort of implication can't be what's meant. $\endgroup$
    – Erick Wong
    Commented Feb 29, 2016 at 17:26
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    $\begingroup$ Be very careful. You want a quantifier on this: $\forall S:[x(S)\implies y(S)]\land \exists S[\lnot (y(S)\implies x(S)]$. I realize that is pendantic, but the logic of $\implies$ is confusing to beginners, and is often taken to mean "can be deduced from," which is the wrong reading. $\endgroup$ Commented Feb 29, 2016 at 17:26
  • $\begingroup$ @ThomasAndrews: It's not pedantic at all. I've already had this issue in the past, which started off simply because I wrote a comment saying "not sure how to write down '$\not\implies$' properly in LaTex". Following commentators suggested that I should strive to avoid using this kind of ambiguous logic. Yours is more accurate, but I felt that using it would make my answer a little too "obfuscated". In any case, I've accepted Erick Wong's comment above, and I will remove this answer, due to the context of the question. Thanks. $\endgroup$ Commented Feb 29, 2016 at 17:29
  • $\begingroup$ But then you should not use the logical symbol $\implies$ at all here, and just write out in text "X implies Y." You are essentially risking a beginner thinking you are writing out propositional logic symbols (especially since you start with "In logic,...") $\endgroup$ Commented Feb 29, 2016 at 17:32
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    $\begingroup$ No, because I am writing text, while you wrote a proposition logic phrase. I am using the colloquial "implies," but propositional logic $\implies$ does not correspond to that collquialism without using quantifiers. This difference between the standard English meaning of implies and the propositional calculus meaning is a constant point of confusion for beginners, and I'm just suggesting to take that in mind and not abuse notation. $\endgroup$ Commented Feb 29, 2016 at 17:42

There are actually theorems in probability that use "strong" and "weak" in their names, for example:

  1. The Law of Large Numbers. The Strong Law of Large Numbers, actually, implies the Weak Law of Large Numbers.
  2. The Central Limit Theorem (CLT). I can't find it online at the moment, but one version (the "strong" version) requires that the random variables are iid and have finite mean and variance. The "weak" version requires that the MGFs of the iid random variables exist, but this assumption can be relaxed to give the "strong" CLT.

Quid's comment on this post uses "stronger" like my example 1. From my comment above: "Usually, when I hear that Theorem B is "stronger than Theorem A" in mathematics, I think of that Theorem B uses less stringent assumptions than Theorem A for a similar result," my example 2 is of this type.

  • $\begingroup$ "[...]I think of that Theorem B uses less stringent assumptions than Theorem A for a similar result." In my perception it is more commonly used when a better a "stronger" conclusion is reached under the same assumptions. Note that your example 1. is of this form, not the one you explain. $\endgroup$
    – quid
    Commented Feb 29, 2016 at 17:27
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    $\begingroup$ @quid See edit. $\endgroup$ Commented Feb 29, 2016 at 17:30

In mathematics I would interpret "stronger" as "more general". In fact, being a result very specific reduces its "power" since it cannot be applied outsite exactly that case. For example, if I show that two continuos functions summed give another continuos function is "weaker" than showing that the sum of any two continuos functions is still a continuos function. This because I can apply the first result only to the two functions I considered, while the second one can be applied to any generic couple of continuos functions (even if the are not actually specified). All this is also the reason of the attention in not losing of generality when you try to prove some result.


Terry Tao (Ask yourself dumb questions – and answer them!):

For instance, given a standard lemma in a subject, you can ask what happens if you delete a hypothesis, or attempt to strengthen the conclusion

To strengthen a conclusion is to say more.

We could have some lemma (or theorem) that says $p \to q$. To attempt to strengthen the conclusion is to see if we can say more than just $q$ so we would try to see if we could say $p \to q_1$ where $q_1$ is some proposition s.t. $q_1 \to q$.

For instance $x=1$ is stronger than $x=0$ or $x=1$. The former implies the latter.

So if we have some assumption that implies the conclusion '$x=0$ or $x=1$', oh let's say, '$x^2 = x$', we would try to see if we could try to strengthen the conclusion to '$x=1$'. We cannot because it is possible that '$x \ne 1$' while '$x^2 = x$' (namely when '$x=0$').

Let's try using a different assumption:

It is true that '$x+1=2$' implies '$x=0$ or $x=1$'. Here, we can strengthen the conclusion to $x=1$.


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