On Huemer’s proof of the minimal free will thesis against determinism

Entradas teóricas., Portada 13 Minutes

Michael Huemer’s book, “Knowledge, Reality, and Value: A Mostly Common Sense Guide to Philosophy,” is a compelling read for those interested in philosophical issues. I have thoroughly studied it over the past few months, and while I generally agree with its ideas, I have a few minor disagreements with it. This book has challenged some of my previous beliefs, particularly regarding animal ethics, and introduced new concepts, such as effective altruism. However, this article will focus on what I believe is one of the book’s biggest errors: Huemer’s argument for free will. For the record, I view the existence of free will in a positive light. I will use the book and his previous article [1] as references for Huemer’s position1.

First, let’s begin by defining some terms (emphasis added):

The minimal free will thesis (MFT) holds that at least some of the time, someone has more than one course of action that he can perform.(1) […] It will be convenient to have a name for the contradictory of MFT. With apologies to compatibilists, I use the label “determinism.” Hereinafter, then, determinism is the thesis that the only thing anyone can ever do is the thing he actually does, where by stipulation, “can” is used in the sense (whatever that is) that is relevant to free will.

[1]

That is, MFT=\neg \det, the negation of determinism. Determinism posits that there is only one possible action at any given time, more details below.

As is well-known, there is something self-refuting about determinism and this idea goes back to Epicurus:

The man who says that all things come to pass by necessity cannot criticize one who denies that all things come to pass by necessity: for he admits that this too happens of necessity.

But this does not necessarily mean that determinism is wrong, it simply says that there is some kind of irrationality. Huemer tries to use this idea to prove the MFT as follows (see §11.4.4 of [2]):

  1. About the free will issue, we should believe only what is most likely to be true. (Premise – presupposed by reasoning)
  2. In general, if S cannot do A, then it is not the case that S should do A. (Premise)
  3. Therefore, with respect to the free will issue, we can believe only what is most likely to be true. (From 1, 2)
  4. If determinism is true, then there is only ever one thing that a person can do; we never have alternative possibilities. (Def. of determinism)
  5. Therefore, if determinism is true, then we believe only what is most likely to be true. (From 3, 4)
  6. I believe in free will. (Premise based on introspection)
  7. Therefore, if determinism is true, then the belief in free will is what is most likely to be true. (From 5, 6)

Stated slightly different (see [1]):

  1. With respect to the free-will issue, we should refrain from believing falsehoods. (premise)
  2. Whatever should be done can be done. (premise)
  3. If determinism is true, then whatever can be done, is done. (premise)
  4. I believe MFT. (premise)
  5. With respect to the free-will issue, we can refrain from believing falsehoods. (from 1,2)
  6. If determinism is true, then with respect to the free will issue, we refrain from believing falsehoods. (from 3,5)
  7. If determinism is true, then MFT is true. (from 6,4)
  8. MFT is true. (from 7)

Let us translate the latter argument with logic symbols and math style, [3], (see also Note 3 of [1] for some ideas):

  1. Premise 1: \forall x\in\{\det,\neg \det\}, \neg x\to S(\neg B(x)).
  2. Premise 2: \forall a \in \mathcal{A}, S(a)\to C(a).
  3. Premise 3: \det\to (\forall a \in \mathcal{A}, C(a)\to D(a)).
  4. Premise 4: D(B(\neg \det)).
  5. \forall x\in\{\det,\neg \det\}, \neg x\to C(\neg B(x)), from steps 1 and 2 with a=\neg B(x).
  6. \det\to (\forall x\in\{\det,\neg \det\}, \neg x\to D(\neg B(x)))
  7. (F_4\land F_6)\to \neg \det. Explanation:
    1. Trivial premise: \forall a \in \mathcal{A}, D(a)\to \neg D(\neg a) with \neg a\in\mathcal{A} (assuming it is closed under \neg up to double negation), i.e., if I do something I do not do the opposite of that something.
    2. \neg D(\neg B(\neg \det)), by the previous step with a=B(\neg \det) and Step 4 using modus ponens. We call this step 4′.
    3. (F_{4'}\land F_6)\land \det\to (\det\land\neg \det), being the last part a contradiction. The result follows from reductio ad absurdum.
  8. \therefore \neg \det, from steps 4,6 and 7.

We have used the following definitions:

  • Predicates, functions: S:=\text{should}, B:=\text{believe}, D:=\text{do}, C:=\text{can}.
  • Other: \mathcal{A}:=\text{set of actions}, F_i:=\text{proposition of the } i\text{-th step}.

More precisely,

S:\mathcal{A}\to \text{Fml}\,,\quad B:\mathcal{X}\to \mathcal{A}\,,\quad D:\mathcal{A}\to \text{Fml}\,,\quad C:\mathcal{A}\to \text{Fml}\,,

being \mathcal{X}:=\{\text{det}, \neg \text{det}\} and \text{Fml} the set of formulae.

Additionally, we have used standard results (as (\forall B) in p.18 of [3]) and some axioms, as the quatifier axioms. For the sake of simplicity, we do not make these technical details explicit.

Thus, the argument is logically valid, F_1\land F_2\land F_3\land F_4 \vdash \neg \text{det}, but is it sound?

Some objections addressed by Huemer

  • Objection #1: Premise (1) begs the question, because, if determinism is true, then it is never the case that a person ‘should’ do anything, because in order for it to be true that S should do A, it must be true both that S can do A and that S can refrain from doing A. Determinism implies that S is never both able to do A and able to refrain from doing A. Therefore, a determinist would obviously reject (1).
  • Objection #2: The argument involves an equivocation, since the “should” in premise (2) is the “should” of morality, while (1) employs the “should” of epistemic rationality.
  • Objection #3: Premise (1) falsely claims that we should believe only what is true. Rather, we should believe only what is justified.
  • Objection #4: (1) is false as used with the epistemic sense of “should,” because people have no control over their beliefs. When a belief is epistemically irrational, there is a sense in which the believer “should not” hold it. However, since people never have a choice about what they believe, this cannot be taken to imply that the believer has it within his power to refrain from holding that belief. To show that people cannot control their beliefs, perform this experiment: try believing that you are a safety pin. You will find that you can’t do it.

My take on these objections (as stated) is that I only consider Objection #3 of some interest. But, at most, these would imply modifying the argument as follows:

  • Premise 1′: \forall x\in\{\det,\neg \det\}, \neg J(x)\to S(\neg B(x)).
  • Step 7′: \det\to J(\neg \det).
  • Step 8′: \therefore \neg\det\lor J(\neg \det).

The new conclusion, as Huemer notes, is also a strong point against determinism. Huemer considers the first premise to be the weakest one (“the first of these is the only one that is at all controversial“), but my critique will be that Premise 2, the ought implies can principle, is not correct in this context.

A different proof based on Premise 2

We are going to give a slightly different proof to understand better the role of Premise 2 in the proof. This premise applied to the free will issue is equivalent to

\forall x\in \{\det,\neg \det\}, S(B(x))\to C(B(x))

which is logically equivalent2 to

\forall x\in \{\det,\neg \det\}, \neg S(B(x))\lor C(B(x)),

i.e., either one can believe x or one should not believe x, because if it were the case that one should believe x and cannot believe x, then this would imply a violation of whatever should be done can be done. The latter, taking all the possible combinations, is equivalent to:

\left(\neg S(B(\det))\land C(B(\neg\det))\right)\lor \left(\neg S(B(\det))\land\neg S(B(\neg\det))\right) \lor \left(C(B(\det)\land C(B( \neg\det)\right) \lor \left(C(B(\det)\land \neg S(B(\neg \det))\right)

More visually, either:

  1. \left(\neg S(B(\det))\land C(B(\neg\det))\right) OR,
  2. \left(\neg S(B(\det))\land\neg S(B(\neg\det))\right) OR,
  3. \left(C(B( \det))\land C(B( \neg\det)\right) OR,
  4. \left(C(B(\det))\land \neg S(B(\neg \det))\right).

Thus, we define F_2=:F_2^1\lor F_2^2 \lor F_2^3\lor F_2^4 where F_2^i is the i-th disjunction of F_2 listed above. Now, we can easily prove \neg \det with the new premises:

  • Premise 1′: \forall x\in\mathcal{X},\neg( x\land \neg S(B(x)), i.e., it is not the case that one should not believe x being x true. This condition is not subject to Objection #3 above.
  • Premise 3′: \neg\left( \det \land C(B(\det))\land C(B(\neg\det))\right)
  • Premise 4′: C(B(\neg\det)), it is a weaker version of Premise 4. It follows from it and the principle “if I do believe x, I can believe x“.

If we assume \det and being \{C_i\}_{i=1}^4 a set of contradictions, then:

  1. F_2^1\land F_1'\land\det \to F_{1}'\land (\neg S(B(\det))\land \det)=:C_1. That is, if the first disjunction of Premise 2 holds, in particular it holds that one should not believe in determinism. This together with determinism is a contradiction assuming Premise 1′ because the second parenthesis is the negation of Premise 1′ for x=\det, determinism holds but one should not believe it.
  2. F_2^2\land F_{1}'\to\left(\det\land\neg\det\right)=:C_2. That is, if the second disjunction of Premise 2 holds, then one should not believe determinism and its negation. But using Premise 1′ this implies the negation of determinism and the negation of the negation of determinism (i.e., determinism), which is a contradiction.
  3. F_2^3\land F_{3}'\land \det\to\left(\neg F_{3}'\land F_{3}'\right)=:C_3, because if one can believe something and the opposite (as the third disjunction of Premise 2 says), determinism cannot be true by definition, Premise 3′.
  4. F_2^4\land F_{3}'\land F_4'\land\det\to \left(\neg\det\land\det\right)=:C_4=C_2, as above but now we use the fact that one can believe in the negation of determinism as an empirical premise, Premise 4′.

So we have shown that each of the four disjunctions of Premise 2 with suitable auxiliary premises, Premise 1′, 3 and 4′, imply a contradiction. Thus, determinism is incompatible with Premise 2 once we accept the auxiliary premises F_1' (should believe what is true), F_4' (empirical premise) and F_3 (definition of determinism). Thus3, either we reject determinism or Premise 2.

The problem of Premise 2

As we have seen, Premise 2 is the key to the proof and in this section, I will argue that this is the problematic hypothesis. Premise 2 is equivalent to \forall a \in \mathcal{A}, \neg C(a)\to \neg S(a), in particular, this implies \forall x \in \mathcal{X}, \neg C(B(x))\to \neg S(B(x)). As stated, this claim is far from obvious. Huemer tries to defend it using several examples (teacher and Bayesian epistemologist). The basic idea:

That is, he who recommends a thing is committed to its being possible to follow his recommendation. If he admits the thing recommended to be impossible, he must withdraw the recommendation.

[1]

Those are examples where a third person is telling others what to do. But it is possible that one should do something although he/she cannot do it because there could be external constraints. Let x\in\mathcal{X} be a proposition of philosophical or scientific interest, and assume that, for some reason, one cannot believe it. If x were true and we want to believe what is true, then S(B(x))\land\neg C(B(x)). Take an extreme example: it can be the case of a person who cannot believe some proposition unless he/she understands a proof of that proposition (because of some kind of skepticism). Let p be some true proposition such that that person cannot understand the proof because of its complexity (e.g., Poincaré conjecture). Thus, as one should believe true propositions and he/she does not understand the proof, \neg C(B(p))\land S(B(p)). So it does not seem the case that we should not believe what we cannot believe. One can imagine more reasons, like a democratic (resp. market) fundamentalist who should believe that some processes can be improved by abandoning democracy (resp. market), but he/she cannot do it because of his/her bias. Following a Moorean style, let us consider the following propositions:

  • \forall ~x\in\mathcal{X}, \tilde{J}(x)\to S(B(x)) , basically, Premise 1,
  • \exists ~x\in\mathcal{X}, \tilde{J}(x)\land \neg C(B(x)) , existence of constraints, at least for one proposition, such that we cannot believe a justified proposition,
  • \forall ~x\in\mathcal{X}, \neg C(B(x))\to \neg S(B(x)) , Premise 2,

where \tilde{J} is either J (justified) or the identity, true. Each of those propositions has some initial plausibility. But they are jointly incompatible (they can’t all be true). Rejecting the second one without further argument is unjustified. There is a gap in the proof. Note also that the quantifier in this sentence is the weakest one, so, ceteris paribus, it increases its plausibility.

A different problem with this proposition in this context is the following. If determinism is true it might be the case that the initial conditions are such that we cannot believe something that is true or that we should believe. More specifically, we can imagine initial conditions, IC_F, such that

\det\land IC_F\to \exists x (\neg x\land D(B(x))),

i.e., x is false but nevertheless one believes in it. More precisely, we are going to see that if Premise 2 were true, together with Premise 1 and Premise 3, then determinism would imply something that is not necessarily true. Indeed, condition \forall x\in\{\det,\neg \det\} in Premise 1 is not necessary and can be substituted by \forall x\in \mathcal{X} for some set of propositions. As Huemer notices:

When we sit down to talk about this issue, or any philosophical issue, there is a tacit assumption that we are all interested in finding out the truth, and we accept this goal as governing the discussion.

[1]

Thus, the new step 6 would be:

  • Step 6′: \det\to (\forall x\in \mathcal{X}, \neg x\to D(\neg B(x))),

i.e., if determinism is true, we do not believe false propositions. But this conclusion seems wrong, as we can imagine initial conditions (as above), I_F, such that \det\land I_F\to \exists x (\neg x\land D(B(x))), i.e., x is false but nevertheless one believes in it, in contradiction with Step 6′. That is,

\det\land I_F\land \left(F_1\land F_2\land F_3\right)\to FB\land\neg FB

where FB:=\exists x (\neg x\land D(B(x))) and F_6'=(\det\to \neg FB). That is, given the initial conditions described above, the premises (note that now F_4 is not needed) and determinism we arrive at a contradiction. Thus, either the premises, determinism or the existence of such initial condition are false.

Hence, although Premise 2 might seem obvious for some examples (especially if a third person is telling you what to do) and true at first glance, it does not seem to be evident for beliefs, especially if determinism is true (what we are trying to elucidate). To be clear, the above reasoning does not prove that the premise is false, but it shows the premise is controversial and it can be rejected unless some new argument is given. That is, given the less uncontroversial premises 1 and 3, what we have shown is that we must reject either \det\land I_F or F_2, but it is not obvious or evident that we should reject the first disjunction. Therefore, accepting F_2 implies that the state of the universe is either a special case of determinism, determinist but all initial conditions ensure that one does not believe falsehoods, or the minimal free will thesis \neg\det (then the proof of \neg\det is trivial once we have Premise 4). But it seems that this step is currently unjustified. Let us insist, if F_2 is true, it is logically clear why that state of the universe is necessary. But the point that I want to stress here is that if we accept Premise 2 (without further justification), then we accept a far from evident restriction of the possible states of the universe. Hence, we might reject the latter, rejecting the Premise 2.

Thus, we have shown that Premise 2 is not evident and it is controversial. If this premise lacks justification, there is a gap in the proof and, therefore, this particular proof of the minimal free will thesis fails.

Footnotes

  1. Huemer recently published a summary of his paper. This was after the article presented here was written.
  2. We say p is provably equivalent to q, if \{\top\}\vdash \left(p\leftrightarrow q\right), being \top a tautology. By the transitivity of the consequence relation ~\vdash~ and the (converse) of the Deduction Theorem, we have \{\top\}\vdash \left(p\to q\right) and \{p\}\vdash q, and that is what we need here. Note also that, using the Soundness Theorem, p\models q and q\models p. Thus, although different as formulae, they are the same proposition.
  3. Obviously, and as the \lor operator indicates, we can reject both.

References

  1. Huemer, Michael. A Proof of Free Will. Available here: http://owl232.net/papers/fwill.htm.
  2. Huemer, Michael. Knowledge, Reality, and Value: A Mostly Common Sense Guide to Philosophy. 2021.
  3. Prestel, Alexander, and Charles N. Delzell. Mathematical logic and model theory: A brief introduction. Springer Science & Business Media, 2011.

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