Gustaf Arrhenius has published a series of impossibility theorems involving ethics. His most recent is The Impossibility of a Satisfactory Population Ethics which basically shows that several intuitive premises yield a stronger version of the repugnant conclusion.

If you know me, you know that I believe that modern ("abstract") algebra can help resolve problems in ethics. This is one example: using some basic algebra, we can get a stronger result than Arrhenius while using weaker axioms.

This is a "standing on the shoulders of giants" type of result: mathematicians have had centuries to trim their axioms to the minimal required set, so once you're able to phrase your question in more standard notation you can quickly arrive at better conclusions. Similarly, the errors in Arrhenius' proof that I've noted in the footnotes are mostly errors of omission that many extremely smart people made, until others pointed out pathological cases where their assumptions were invalid.

### Assumptions

We assume that it's possible to have lives that are worth living ("positive" welfare), lives not worth living ("negative" welfare) and ones on the margin ("neutral" welfare). Arrhenius doesn't specify what the relationship is between "positive" and "negative" welfare, but I think there's a very intuitive answer: they cancel each other out. Just as $(+1) + (-1) = 0$, a world with a person of $+1$ utility and one with $-1$ utility is equivalent to a world with people at the neutral level.

^{1}

We continue the analogy with addition by writing $Z=X+Y$ if $Z$ is the union of two populations $X$ and $Y$. Just as with normal addition, we assume that $X+Y$ is always defined

^{2}and that we can move parentheses around however we want, i.e. $(X+Y)+Z=X+(Y+Z)$. Lastly, I'm going to assume that the order in which you add people doesn't matter, i.e. $X+Y=Y+X$.

^{3}I will finish the analogy with addition by specifying that welfare is isomorphic to the integers.

^{4}

(The above is just a long-winded way of saying that population ethics is isomorphic to the free abelian group on $\mathbb Z$.)

Also, for simplicity, I will write $nX$ for $\underbrace{X+\dots+X}_{n\ times}$.

^{5}

Lastly, we need to define our ordering. I'll use the notation that $X\leq Y$ means "Population $X$ is morally worse than population $Y$" and require that $\leq$ is a quasi-order, i.e. $X\leq X$ and $X\leq Y, Y\leq Z$ implies that $X\leq Z$. Notably, this does not require us to believe that populations are totally ordered, i.e. there may be cases where we aren't sure which population is better.

The major controversial assumption we need from Arrhenius is what he calls "non-elitism": for any $X,Y$ with $X-1>Y$ there is an $n>0$ such that for any population $D$ consisting of people with welfare levels between $X$ and $Y$: $(n+1)(X-1)+D\geq X+nY+D$. In less formal terms, this is basically saying that there are no "infinitely good" welfare levels.

### Claim

We claim that any group following the above axioms results in:

The Very Repugnant Conclusion: For any perfectly equal population

with very high positive welfare, and for any number of lives with very

negative welfare, there is a population consisting of the lives with negative welfare and lives with very low positive welfare which is better than the high welfare population, all things being equal.

### Unused Assumptions

The following are assumptions Arrhenius makes which are unused. (Note: these are verbatim quotes from his paper, unlike the other assumptions.)

(Exercise for the advanced reader: figure out which of these also follow from the assumptions we did use.)

*The Egalitarian Dominance Condition*: If population A is a perfectly

equal population of the same size as population B, and every person in

A has higher welfare than every person in B, then A is better than B,

other things being equal.*The General Non-Extreme Priority Condition*: There is a number n

of lives such that for any population X, and any welfare level A, a

population consisting of the X-lives, n lives with very high welfare, and

one life with welfare A, is at least as good as a population consisting

of the X-lives, n lives with very low positive welfare, and one life with

welfare slightly above A, other things being equal.*The Weak Non-Sadism Condition:*There is a negative welfare level and

a number of lives at this level such that an addition of any number of

people with positive welfare is at least as good as an addition of the

lives with negative welfare, other things being equal.

### Proof

**Lemma**

First we prove a lemma: what Arrhenius calls "Condition $\beta$" and what mathematicians would refer to as a proof that our group is Archimedean. This means that for any $X,Y>0$ there is an $n$ such that $nX\geq Y$.

Basically we just observe that the "non-elitism" condition makes a simple induction. Starting from the premise that $(n+1)(X-1)+D\geq X+nY+D$, let $Y, D=0$, giving us that $(n+1)(X-1)\geq X$, i.e. $X$ is Archimedean with respect to $X-1$. Continuing the induction we find that $X$ is Archimedean with respect to $X-k$, completing the proof.

^{6,7}

**Theorem**

First, let me give a formal definition of the "Very Repugnant Conclusion": For any high level of welfare $H$, low positive level of welfare $L$ and negative level of welfare $-N$ and population sizes $c_{H},c_{N}$ there is some $c_{L}$ such that $c_{L}\cdot L+c_{N}\cdot(-N)\geq c_{H}H$.

To prove our claim: we know there is some $k_{1}$ such that

$$k_{1}\cdot L\geq c_{H}\cdot H\label{ref1}$$

because of our lemma. Because it's a group, we know that $(N+-N)+L=L$ and moreover $(c_{N}N+c_{N}\cdot-N)+L=L$. Substituting this into (1) yields

$$k_{1}\left[\left(c_{N}N+c_{N}\cdot-N\right)+L\right]\geq c_{H}H\label{ref2}$$

Expanding the left hand side of (2) we get

$$k_{1}c_{N}N+k_{1}c_{N}\cdot(-N)+k_{1}L\label{ref3}$$

By our lemma there is some $k_{2}$ such that $k_{2}L+D\geq k_{1}c_{N}N+D$; letting $D=k_{1}c_{N}(-N)+k_{1}L$ and using transitivity we get that

$$k_{2}L+k_{1}c_{N}(-N)+k_{1}L\geq c_{H}H$$

Rewriting terms leaves us with

$$\left(k_{1}+k_{2}\right)L+k_{1}c_{N}(-N)\geq c_{H}H$$

or

$$c_L L+c_{N'}(-N)\geq c_{H}H$$

$\blacksquare$

### Comments

I don't know that this shorter proof is much more convincing than Arrhenius' - my guess is that the people who disagree with an assumption are those who take a "person-affecting" view or otherwise object to the entire premise of the theorem. I would though say that:

- None of the math I've used is beyond the average high-school student. It's just making the "algebra can be about things other than numbers" leap which is hard.
- While abstract algebraic notation can be intimidating, it's relevant to realize that using it makes you more concise. (To the extent that a 26-page paper can be rewritten into a two-page blog post.)
- Because we can be more concise and use standard terminology, it shines a light on what is really the controversial assumption: Non-Elitism.
- Similarly, because we use standard concepts it's easier to see missing assumptions (e.g. I didn't realize that Arrhenius was missing a closure axiom until I tried to cast it in group theory terms).

### Acknowledgements

I would like to thank Prof. Arrhenius for the idea, and Nick Beckstead for talking about it with me.

### Footnotes

- Formally, for each $X$ there is some $-X$ such that for all $Y$, $X+(-X)+Y=Y$.
- This isn't an explicit assumption in Arrhenius, but it's implicitly assumed just about everywhere
- This arguably is controversial so I'll point out that commutativity isn't really required, but since it keeps the proof a lot shorter and most people will accept it, I'll keep the assumption
- Arrhenius "proves" that welfare is order-isomorphic to $\mathbb Z$ incorrectly, so I'll just assume it instead of attempting to derive it from others. If you prefer, you can take his "Discreteness" axiom, add in assumptions that welfare is totally ordered and has no least or greatest element and you'll get the same thing.
- Which is just to say that since it's an abelian group it's also a $\mathbb Z$-module.
- Nick Beckstead thought that some people might not like using the neutral level like this, so I'll point out that you can use an alternative proof at the expense of an additional axiom. If you assume non-sadism, then you can find that $X+nY\geq X$ and therefore transitively $(n+1)(X-1)\geq X$.
- This is somewhat misleading: we've only shown that the group is archimedean for totally equitable populations. That's all we need though.

Dear XODARAP,

ReplyDeleteUnless I am missing something obvious, I believe you have completely missed the point of Arrhenius's paper. First you say you will use variables X, Y, to represent populations. But then you go on to use them to represent amounts of welfare. That would be ok if we assume that the total value of a population = the sum of the welfare of the individuals in it. In other words, it would be ok if we assumed Total Utilitarianism. But the very repugnant conclusion is entirely trivial on that assumption.

The point is that people in population ethics are looking for theories where the value of a world is different from the sum of the value of the lives in it. Arrhenius's theorem is meant to prove that we still get the very repugnant conclusion, given the assumptions he makes. That's why he needs the assumptions you say he doesn't need.

A different way of saying what I'm trying to say is that your "for simplicity, I will write nX for X+⋯+X (n times)" makes no sense. Mary+Mary+Mary+Mary doesn't equal anything.

A third way of saying the same thing is to ask: what do your variables X, Y, represent?

All the best,

Knut

Hey Knut,

DeleteThank you for your comment - I agree that this notation can be confusing, and I should've explained this point more clearly.

As I mentioned, $Z=X+Y$ means $Z$ is the union of two populations $X$ and $Y$. So Mary+Mary+Mary+Mary would mean "A population with four people in it, each who have Mary's quality of life". It would _not_ mean "A population consisting of one person whose quality of life is four times Mary's".

So I don't think this makes an assumption about being total utilitarian - plenty of groups meet the assumptions of my post, but are not (R,+).