# Different approaches to proofs that "are the same"?

This question (and answers) on MSE got me thinking on simple examples of different ways of proving the same (hopefully somewhat interesting) result, as examples to be discussed on difference in clarity or whatever. This for use in a first course in discrete mathematics, for students in the middle of the typical college calculus sequence.

Particularly interesting would be proofs that use different techniques (contradiction, contrapositive) but that one would consider be "the same proof, written up differently".

• I wrote up a sample lecture on MO and re-mentioned it on MESE. The proofs use different techniques, but knowing, e.g., that every whole number can be written as even or odd essentially hinges on mathematical induction (etc). I'm not sure if the result (i.e. that the product of consecutive naturals is always even) is "interesting" and it is number theory rather than discrete mathematics. But I did write up a version to be submitted for publication, so let me know if this fits the bill... Feb 2 '16 at 22:12
• @BenjaminDickman, just too bad that crossposting is frowned upon. Feb 2 '16 at 22:53

A source of simple example is in basic set theory. Let $A,B,C$ be sets and $f:A\to B$, $g:B\to C$ maps. Consider assertions of the kind "if $f$ and $g$ are one-to-one, then $g\circ f$ is one-to-one".

You can prove it directly using this classical definition "$f$ is one-to-one when for all $x,y\in A, f(x)=f(y) \implies x=y$". Assume $f$ and $g$ are one-to-one and let $x,y\in A$ be such that $g(f(x))=g(f(y))$. Since $g$ is one-to-one, we have $f(x)=f(y)$. Since $f$ is one-to-one, we thus have $x=y$. We have proved that $g\circ f$ is one-to-one.

You can reformulate using the contrapositive definition "$f$ is one-to-one when for all $x,y\in A, x\neq y\implies f(x)\neq f(y)$". Assume $f$ and $g$ are one-to-one and let $x\neq y\in A$. Since $f$ is one-to-one, $f(x)\neq f(y)$. Then since $g$ is one-to-one, $g(f(x))\neq g(f(y))$.

You can prove by contradiction: assume $f$ and $g$ are one-to-one and that there are $x\neq y\in A$ such that $g(f(x))=g(f(y))$. Either $f(x)= f(y)$, contradicting the assumption that $f$ is one-to-one, or $f(x)\neq f(y)$ and $g(f(x))=g(f(y))$ contradicts the assumption that $g$ is one-to-one.

You can use a contrapositive: assume $g\circ f$ is not one-to-one, and prove that either $f$ or $g$ is not one-to-one (as above).

There are probably others... and I would say that all these proofs are the same under different guises. I believe that there exist no proofs of this fact that I would not qualify as essentially the same as the above, but I would be delighted to be proven wrong in a comment!

• You won't be proven wrong; all proofs have the same moral core. (Now I better run off before someone asks me to rigorously define "moral".) On another note, I think students should develop an intuitive feeling of when something requires proof by contradiction, namely $( \neg P \rightarrow \bot ) \vdash P$. Not only can constructive proofs can be converted into algorithms, another good reason is that any proof can be trivially transformed into a proof by contradiction, so if one cannot prove something "directly" the next step is to try contradiction, but one has to know when to stop. Feb 3 '16 at 5:19

The Vandermonde binomial identity.

The Wikipedia page gives three proofs; an algebraic one (symbol pushing), a combinatorial one (counting in two different ways), and a geometric one (where a certain family of paths is decomposed recursively). All three proofs are short, simple, and insightful, and they complement each other.

One important example is the equivalence of standard bottom-up induction and top-down induction. It is interesting that bottom-up induction does not extend immediately to ordinals (since successors and limits must be handled differently), while top-down induction extends without any change. On the other hand, top-down induction is usually easier to learn without error, especially in the form of structural induction (see this explanation).

Closely related is the equivalence of induction and well-ordering. Here there is an interesting disparity, because in some cases considering the smallest counter-example with sizes from some well-order yields much easier proofs, partly because it is essentially induction wrapped in a non-constructive proof by contradiction, which is in some sense strictly stronger than constructive induction, namely when in a constructive logic. This may seem irrelevant, but in many cases it is far easier to see a proof by well-ordering (also called extremal principle) than to see the equivalent proof by induction (see this short list on AOPS.

For analysis in particular, it's interesting to look at the proofs of basic results such as that continuous functions on a compact set are bounded and attains its extrema, and are uniformly continuous. Because of the equivalent notions of compactness for metric spaces, it is interesting to look at the resulting differences in proofs. One would notice though that the core of each proof is pretty much the same for dimension at least 2. However, in 1 dimension there are rather different proofs that I discovered long time ago that do not extend (at least I don't see how), which I give as an example below.

There are many ways to precisely quantify sameness of proofs. One way that is good for students to learn is to see exactly where each part of the theorem and which axioms are used in the proof, and also count the number of times each of them is used. Two proofs are essentially the same when they are minimal in that respect and use each part the same number of times. This can be done at any level of detail, but usually we do not include any of the logical axioms except for double negation elimination (equivalently proof by contradiction in the form "$$\neg P \rightarrow \bot \vdash P$$"), and often we work over a specific axiomatization such as that of the real numbers (second-order completeness). Here is an example: $$\def\rr{\mathbb{R}}$$ $$\def\less{\smallsetminus}$$

Any continuous function $$f : [0,1] \to \rr$$ is bounded on $$[0,1]$$.

Proof [with reason on the right of each line]:

Take any continuous function $$f : [0,1] \to \rr$$. [logically necessary]

Let $$S = \Big\{ p : p \in [0,1] \land \text{f is bounded on [0,p]} \Big\}$$. [by comprehension]

Then $$0 \in S$$ because $$\forall x \in [0,0]\ ( |f(x)| \le |f(0)| )$$. [this uses the left endpoint of $$[0,1]$$]

Then $$S$$ is non-empty. [by definition of non-empty]

Also, for any $$p \in S$$, we have $$p \le 1$$ by construction of $$S$$. [this uses the right endpoint of $$[0,1]$$]

Thus $$S$$ has a least upper bound $$c$$ in $$\rr$$. [by completeness of real numbers]

Firstly $$c \ge 0$$ because $$0 \in S$$. [this uses that $$c$$ is an upper bound]

Secondly $$c \le 1$$ because $$1$$ is an upper bound for $$S$$ in $$\rr$$. [this uses that $$c$$ is the least one]

Let $$δ > 0$$ such that, for any $$x \in [0,1]$$, if $$|x-c| \le δ$$ then $$|f(x)-f(c)| \le 1$$. [this uses continuity but is much weaker (*)]

Thus $$f$$ is bounded on $$[0,1] \cap [c-δ,c+δ]$$ by $$|f(c)|+1$$. [by triangle inequality]

Also $$f$$ is bounded on $$[0,c-δ]$$ by definition of $$S$$. [this is why we defined $$S$$ like that]

Thus $$f$$ is bounded on $$[0,1] \cap [0,c+δ]$$. [by basic properties of reals]

Thus $$\min(1,c+δ) \in S$$ by definition of S. [in the reverse direction]

Thus $$\min(1,c+δ) \le c$$ and hence $$1 \le c$$. [by basic properties of reals]

Therefore $$f$$ is bounded on $$[0,1]$$ since $$[0,1] \subseteq [0,c+δ]$$.

(*) This is a clear example of an important point that students should learn to identify and understand. Continuity of $$f$$ is essentially an assertion of the form "$$\forall ε > 0\ ( \text{f is ε-continuous} )$$", and here all we used was $$1$$-continuity of $$f$$, and $$1$$ was arbitrary except that it was more than $$0$$. Students with a proper foundation in logic would not only be able to prove the theorem but also realize via this observation that we can strengthen it to:

Given any $$ε > 0$$, any $$ε$$-continuous function $$f : [0,1] \to \rr$$ is bounded on $$[0,1]$$.

Notice that in the above proof every part of the theorem statement has been used exactly once. It is instructive to check that dropping any single part, such as continuity of $$f$$, or $$[0,1]$$ being closed on each end, results in a false assertion. With a suitable notion of shortness, this implies that the proof is more or less minimal over the standard axiomatization of the reals, which anyone reading the proof would expect.

Note however that theorems and proofs do not live in isolation, and so one can consider any particular body $$B$$ of mathematics as a whole, and ask what is the shortest proof of all the theorems in $$B$$. This minimization tends to have all the key theorems factored out, much like in Lempel-Ziv compression. What we are minimizing is essentially the Kolmogorov complexity of $$B$$ with respect to the underlying formal system (such as ZFC).

For any collection $$C$$ of theorems over an existing body $$B$$ where $$B$$ has a minimal proof, one can look at the proof of all the theorems in $$C$$ over $$B$$. Two proofs are then reasonably considered identical if they use the same axioms and the same proven sentences in $$B$$. To make sure this notion coincides with our intuition, we can require that a sentence must be labelled before it can be used as a theorem, so that only major results will be labelled in a minimal proof of $$B$$.

If the two proofs $$P,Q$$ of $$C$$ over $$B$$ are not considered identical according to the above, we can measure the difference via some reasonable proof distance, such as one that I just thought of that takes into account differences between theorems used. First let $$S,T$$ be the collection of all theorems in $$B$$ that $$P,Q$$ respectively depend on directly. Then let $$R$$ be a minimal proof of $$S \cup T$$. (Give each equal weight if there is more than one.) Then the proof distance between $$P,Q$$ is the edit distance between theorems and axioms used in $$P$$ and $$Q$$ ignoring order, plus the total length of the segments of $$R$$ that are used only for one of $$S,T$$. This definition ensures that two proofs that have the same essential ingredients have small proof distance, and proof distance is only slightly affected if the same theorem is proven and used under different names, since minimizing the proof of $$S \cup T$$ automatically finds the common ingredients.

Likewise, and in fact more easily, we can define similarity in mathematical content of theorems. Given any two collections $$A,B$$ of theorems, let $$R$$ be a minimal proof of $$A \cup B$$. Then the content distance between $$A$$ and $$B$$ is just the total length of the segments of $$R$$ that are used only for one of $$A,B$$.

This is quite technical, but I believe it quite accurately captures what we mean when we say things like "Compactness and completeness are essentially equivalent theorems about first-order logic.". Of course all true assertions are equivalent, including "$$0 = 0$$.", but we all know that the statement is trying to say that the easiest way to prove both compactness and completeness is to prove one and then derive the other as a corollary. Indeed if we let $$A$$ be the singleton containing "compactness" and $$B$$ be the singleton containing "completeness", then since there is a short proof of $$A$$ over $$B$$ and of $$B$$ over $$A$$, the shortest proof of $$A \cup B$$ is thus only a few lines longer than the common ingredients of the individual proofs, and thus there is negligible content difference between compactness and completeness, as expected.

A comment on the question you link to makes a valid point: it's important to define what you mean by "different ways of proving".

From the logical standpoint, the difference depends on the axioms or first principles from which the theorem is deduced. Two proofs that argue from the same axioms are logically equivalent. If one wants to prove that $p \Rightarrow q,$ it may be possible to demonstrate this directly, or one may show that $!p \Rightarrow !q$, which is a proof by contraposition. Logically, these are equivalent proofs, so they may not necessarily be thought of as "different."

If you were to deduce a statement from a set of axioms, then from an independent set of axioms, that would certainly constitute a different proof by any definition. This is certainly possible in many cases, for example showing that a certain object does not satisfy a set of axioms (group, ring, etc.).

For example, any statement of set theory that can be deduced from the well-ordered principle can also be deduced from the principle of induction, because it can be shown that those two axioms are logically equivalent.

For this reason, I tend to think of axioms as sort of a "logical basis," where basis here is meant in the algebraic sense of a spanning set, in this case the theorems are the span of the axioms, if you will. As you search for examples, try to identify when the proofs follow from the same "basis" set, and when they do not.

• For didactical purposes (first encounter with more rigorous proofs), the axioms are certainly important, but I don't want to dwell on "this two sets of axioms are equivalent" (no time for that, unfortunately). Feb 2 '16 at 19:52
• "Two proofs that argue from the same axioms are logically equivalent": how do you define logically equivalent proofs? I have the feeling that defining intrinsically different proofs is both subtle and meaningful (and after following links, I see that there is much to say: mathoverflow.net/questions/3776/…) Feb 2 '16 at 21:24
• @BenoîtKloeckner, that is a can of worms I prefer leaving unopened. It is enough for my purposes that the proofs look different. Feb 3 '16 at 0:26
• You don't want to only look at the axioms used. Say you work in ZFC, the standard set theory in which almost all modern mathematics can be carried out after a suitable translation. In ZFC there are only a few axiom schemas. You can't really say that you distinguish proofs that use exactly the same ideas but just use different instances of one of the axiom schemas. But there are only finitely many axiom schemas, which would mean that in your reckoning there are only finitely many different proofs... However, I do know what you're trying to say, and I suggest one distance measure my answer. Feb 3 '16 at 13:23
• @vonbrand Yes, there are a significant amount of philosophical considerations at play here- while I stand by my answer and I think that it points out important considerations (which was my intention), I hope that the other answers provide more concrete help. Feb 3 '16 at 14:40

Euclid's proof that there are infinitely many primes is typically presented as a proof by contradiction but sometimes as a positive proof which provides a recipe for constructing primes not in a given set of primes (which yields the infinitude as an immediate corollary). The argument is essentially the same in both cases, but the two ways of presenting the proof have very different flavors.