Zombies Schmonceivability and the Mirroring Objection All EscapeRoutes
Zombies, Schmonceivability and the Mirroring Objection: All Escape-Routes Barred Doug Campbell Department of Philosophy
Background
Materialism • • P – a conjunction of the actually obtaining microphysical facts. T – a ‘totality operator’ (or ‘that’s all’ clause). PT – says that no facts obtain beyond those supervenient on P. Q – a conjunction of all actually obtaining facts, including the phenomenal facts. Materialism: • It is thesis □(PT→Q) – that any ‘PT-world’ must also be a ‘Q-world’ – that the phenomenal facts (and all other facts) supervene metaphysically on the bare microphysical facts – that ‘a minimal physical duplicate of the actual world is a duplicate simpliciter’
Materialism Minds and Qualia If materialism is true then God thereby automatically fixes all the actually obtaining facts, including the qualitative facts. (All the actually obtaining facts come ‘for free’ with PT. ) (Here we conveniently turn a blind eye to the fact that God’s existence is probably incompatible with materialism!) God fixes all the microphysical facts, and then says “that’s all”. I. e. , God makes PT true.
What it is like to be a normal person riding a mountain bike
What it is like to be a philosophical zombie riding a mountain bike (Except it is not even black. What is like to be a philosophical zombie matches what it is like to be in a dreamless sleep, or in a coma, or dead. )
What a philosophical zombie does not look like
What a philosophical zombie does look like
A Philosophical Zombie From the outside she appears to be a perfectly normal person. From a behavioral, psychological , or neurophysiological perspectives she is perfectly normal. But she has no inner mental life. She has no subjective experiences. There is “no one at home”. She will claim to be conscious, just like a normal person. But really, she isn’t.
Austere Conceivability • p is austerely conceivable iff p is free of any a priori detectable contradictions. – I. e. , p is austerely conceivable iff no one, no matter how highly rational they might be, could ever find a contradiction in p by a process of mere rational reflection. • Notation: ◇cp represents the claim that p is austerely conceivable. • Henceforth by ‘conceivable’ I will mean ‘austerely conceivable’ unless I explicitly say otherwise. • In fancy symbols:
The principle that Conceivability Entails Possibility (CEP)
The zombie argument against materialism The conceivability of a zombie world (where PT is true but Q is not) CEP Logical truth The denial of materialism
The usual materialist responses Attack Z 1. Zombies are not austerely conceivable � � Attack Z 2: conceivability is not a reliable guide to possibility
Our response: the mirror argument The conceivability of a world where PT is true and Q is true too. CEP Logical truth Materialism
Why is M 3 a logical truth? • M 3 says, in effect, that if there is some PT world that is a Q world, then every PT world is a Q world. • This is true because of the meaning of the T operator. – A PT world is a world where all the facts supervene on the P facts. – Thus any two PT world must be exactly alike. – Hence if some PT world is a Q world, then every PT world will be a Q world (just as M 3 says).
Chalmers dilemma, short version So it seems he must deny M 1. And he can’t deny M 2 without undermining CEP, which he relies on to justify M 2 But he can’t deny M 3, since it is a tautology Chalmers can’t accept M 4, since it contradicts his own conclusion. Hence he must deny M 1, M 2, or M 3
Why Chalmers can’t easily deny M 1 • In denying M 1, he would be saying PT∧Q is inconceivable. • That is, he would be saying that PT∧Q entails some a priori detectable contradiction.
The inconceivability argument against materialism (IAAM) IEI is about as uncontroversial as any modal principle could be. It is implied by one of the most fundamental rules of modal logic, the Necessitation Rule The denial of M 1 IEI A tautology The denial of materialism
Why is A 3 a tautology? • Consider a PT-world, w (a possible world that is a minimal physical duplicate of our actual world). – Such a w certainly exists. If physicalism is true then w will be identical to the actual world, while if physicalism is false then w will be a bare physical duplicate of the actual world from which non-physical things have been subtracted. • A 3’s antecedent says, in effect, that no PT-world is a Q-world. Assume this is true. • Then it follows that w is not a Q-world. • Thus there is at least one world, w, that is a PT-world but not a Q-world. • Thus it is not the case that every PT-world is a Q-world. • This is what A 3’s consequent says. • And so, assuming A 3’s antecedent is true, its consequent is true too. • Thus, A 3 itself is true. Q. E. D.
In short, if Chalmers is in a position to show that M 1 is false, then he is in a position to provide an absolutely logically watertight, IAAM-based deductive refutation of materialism. Thus he doesn’t need the zombie argument in the first place. The zombie argument is redundant.
The situation is symmetrical
Chalmers’ response • “It may be prima facie negatively conceivable that materialism is true about consciousness, but it is not obviously conceivable in any stronger sense. Many people have noted that it is very hard to imagine that consciousness is a physical process. I do not think that this unimaginability is so obvious that it should be used as a premise in an argument against materialism, but likewise the imaginability claim cannot be used as a premise either. ” (Chalmers, 2010, p. 180)
Chalmers’ response (recent email) (He doesn’t do capitals in emails!) • i suppose i think our initial intuitions favor the inconceivability claim, so based on those it's at least 75% plausible, say. that does allow me to assign 75% credence to the falsity of materialism even without bringing in the zombie argument. however, (i) this argument is dialectically weak as the premise is so easy to deny, and (ii) the zombie argument allows a much stronger conclusion -- since i think it's much more than 75% plausible that zombies are conceivable, leading to much greater credence that materialism is false. for both of these reasons ZA isn't redundant.
Chalmers’ response ✓(99. 9%) ✓ (75%) ✗ (75%)
In our new article we attempt to systematically examine all Chalmers’ options, including the option he actually endorses, and show that all carry great costs.
Chalmers Options The Anti-M 1 Route: he denies M 1 The anti-M 2 route: he denies M 2. But he can’t deny M 3, since it is a tautology Chalmers can’t accept M 4, since it contradicts his own conclusion. Hence he must deny M 1, M 2, or M 3
The anti-M 1 route • If M 1 is false, then there is some valid sequence of logical steps, S, by which a contradiction can be deduced from PT∧Q.
The anti-M 1 route, continued • The anti-M 1 route can be divided into three sub-routes: 1. The mysterian route. – Chalmers say he has not been able to find S. (S remains ‘mysterious’. ) 2. The containment route. – Chalmers says that S contains the zombie argument as a component. 3. The buttressing route. – Chalmers says he has found S, and that it doesn’t contain the zombie argument as a component, but then claims that logical redundancy is not a bad thing. After all, it is better for an antiphysicalist to have two arguments up her sleeve than only one, since this enables each argument to be used to back-up, or ‘buttress’, the other.
The mysterian route, critiqued • Cost A: admitting that the anti-physicalist has the pressing burden of demonstrating S. – Chalmers is caught in an embarrassing position if he insists that S exists, but cannot find it. • Cost B: admitting that the mysterian route is inherently unstable, and that whoever endorses it must ultimately also endorse some other reply to the mirroring objection too. – As soon as Chalmers find S, then S will be ‘mysterious’ no more, and so he will no longer be able to avail himself of the mysterian route.
Another problem with the mysterian route • Under the mysterian route, Chalmers claims that M 1 is false (and thus that PT∧Q is contradictory) even though he can’t show us the contradiction in question. • But this allows the materialist to turn the tables on Chalmers. – The materialist can pull the same trick, by claiming that Chalmers’ premise, Z 1 is false (i. e. , that PT∧¬Q is contradictory). – The materialist needn’t feel any need to demonstrate the contradiction in question. – Sauce for the materialist’s goose is sauce for Chalmers’ gander.
• To overcome make the mysterian route work Chalmers would need to break the symmetry between his position and that of his physicalist foe by showing that there is a stronger case for thinking M 1 is false (and thus that PT∧Q is contradictory) than for thinking Z 1 is false (and thus that PT∧¬Q is contradictory).
Option 1. • Might Chalmers rely on the fact that no contradiction has yet been found in PT∧¬Q in order to show that, probably, no such contradiction exists, and thus that Z 1 is probably true? • No, because the mysterian route requires him to admit that no contradiction has yet been found in PT∧Q either. By parity of reasoning he should conclude that M 1 is likewise probably true.
Option 2. • Might he try to break the symmetry between Z 1 and M 1 by appealing to asymmetry of contradiction-hunting effort? – Might he argue that philosophers have searched more assiduously for a contradiction in PT∧¬Q than in PT∧Q, and that this gives us stronger reasons for thinking the former proposition is indeed contradiction-free (and thus that Z 1 is true) than for thinking the latter proposition is contradictionfree (and thus that M 1 is true)? • No. – This would the principle that our credence in a proposition being contradiction-free should be proportionate to the amount of unsuccessful effort we have put into finding a contradiction in it. This principle is not plausible. – If anti-physicalists were able to find a contradiction in PT∧Q then they would be in a position to offer a knockdown, IAAMbased refutation of physicalism. • This being so, they would be very poor philosophers indeed if they had not searched long and hard for a contradiction in PT∧Q.
Option 3. • To ‘positively conceive’ of a situation is to model it in one’s imagination, and thereby construct a kind of (low grade) ‘existence proof’ that it is possible and therefore contradiction-free. • Might Chalmers try to break the symmetry between Z 1 and M 1 by arguing that we can positively conceive PT∧¬Q but not PT∧Q? • He could make this claim, but it is very difficult to see why a physicalist should accept it. – To the contrary, any physicalist worth her salt should respond by insisting that PT∧Q is every bit as positively conceivable as PT∧¬Q. The physicalist, after all, thinks PT∧Q is actually true. – As far as she is concerned, the act of positively conceiving PT∧Q is identical to the act of conceiving actually obtaining states of affairs. – It is not the mere contemplation of a counterfactual, as is the antiphysicalist’s act of conceiving of PT∧¬Q. – No imaginative leap is required! – The physicalist positively conceives of PT∧Q all the time, so to speak, in everyday life.
Option 4. • Might Chalmers try to tilt the balance for Z 1 and against M 1 by appealing to the idea that it is somehow easier to imagine PT∧¬Q being true (e. g. , by imagining zombies) than to imagine PT∧Q being true (e. g. , by imagining conscious states to be identical to physical or functional brain states)? • This approach is untenable for two reasons. 1. Physicalists will not accept the premise, since they find it perfectly easy to imagine PT∧Q being true. (After all, they think it is true. ) 2. It relies on the dubious principle that the ease with which we can imagine a proposition’s being true is a reliable guide to its chances of being contradictionfree.
Option 5. • Might Chalmers break the symmetry between Z 1 and M 1 by providing a full deductive proof that Z 1 is true? • Maybe. • This would work only if the proof of Z 1 was “asymmetrical”, in the sense of not being susceptible to being generalized into a sound proof of M 1. • Cost C: admitting that it is the anti-physicalist’s burden to provide an asymmetric Z 1 Proof, and admitting that extant versions of the zombie argument, which do not incorporate such a proof, are fallacious because they can be mirrored.
So: the mysterian route has three prohibitive costs • Cost A: admitting that the anti-physicalist has the pressing burden of demonstrating S. • Cost B: admitting that the mysterian route is inherently unstable, and that whoever endorses it must ultimately also endorse some other reply to the mirroring objection too. • Cost C: admitting that it is the anti-physicalist’s burden to provide an asymmetric Z 1 Proof, and admitting that extant versions of the zombie argument, which do not incorporate such a proof, are fallacious because they can be mirrored.
The containment route • A car’s carburetor has the job of helping to power the car. But the car is powered only by its engine. Does it follow that the carburetor is redundant? • No! The engine contains the carburetor as an indispensable part. The engine powers the car partly by virtue of the carburetor mixing air with fuel. • Under the containment route, Chalmers would hold that the zombie argument is related to S+IAAM much as the carburetor is related to the engine: it is not made redundant by S+IAAM, because in using S+IAAM to refute materialism the anti-materialist is (among other things) relying on the premises and conclusion of zombie argument. The zombie argument is an indispensable part of S+IAAM.
How could S+IAAM have the zombie argument as a component?
However, the situation is symmetrical
• Thus in order to make the containment route work, Chalmers must prove D 1, D 2 and D 3, while also showing that at least one of these proofs is ”asymmetrical”, in the sense that E 1, E 2 and E 3 cannot be proved by the same methods. • Since both Z 3 and M 3 are tautological, so too are D 3 and E 3.
Thus the containment route carries the following cost: • Cost D: admitting that it is the antiphysicalist’s burden, first, to produce both a D 1 proof and a D 2 proof, and second, to show that at least one of these two proofs is asymmetrical; and also admitting that extant versions of the zombie argument, which do not incorporate such proofs, are fallacious because they can be mirrored.
The buttressing route • Cost E: admitting that it is the anti-physicalist’s burden to immediately explain how a contradiction can be deduced from PT∧Q by following a sequence of steps that do not include inferring the denial of physicalism from Z 1, Z 2 and Z 3.
The specter of logical redundancy • Suppose Chalmers can in fact show why PT∧Q is contradictory. This will then enable him to offer an IAAM-based refutation of materialism. • This suggests that if Chalmers can show PT∧Q is contradictory, then the zombie argument serves no useful purpose. • There seems to be only one way in which Chalmers might respond to this objection— namely, by arguing that the zombie argument usefully buttresses and supports the IAAM-based argument against materialism.
Buttressing or logical redundancy? • G an H are both arguments for C. • But G’s premises are susceptible to certain objections. In order to defend G’s premises from these objections, it is necessary to invoke H’s premises. • Does G provide any sound reason to believe C over and above the reason that H provides?
S (a demonstration that PT∧Q is contradictory) Mirror argument attack
The buttressing route • Cost F: admitting that the zombie argument is logically redundant.
The anti-M 2 route The Anti-M 1 Route: he denies M 1 The anti-M 2 route: he denies M 2. But he can’t deny M 3, since it is a tautology Chalmers can’t accept M 4, since it contradicts his own conclusion. Hence he must deny M 1, M 2, or M 3
The anti-M 2 route • The anti-M 2 route can be divided into two sub -routes. 1. The pro-CEP route. Chalmers upholds CEP. 2. The anti-CEP route. Chalmers opts to repudiate CEP.
The pro-CEP route • If Chalmers is going to uphold CEP while denying M 2 then he must hold that CEP has a ‘restriction clause’ that excludes PT∧Q from its scope. • At the same time, he must hold that CEP does not have a restriction clause that excludes PT∧¬Q from its scope—since otherwise he wouldn’t be able to use CEP to justify Z 2.
CEP restriction clauses • Both the below are plausibly conceivable but impossible, and hence counterexample to CEP. 1. Water is not H 2 O. 2. Goldbach’s conjecture is false—i. e. , there does exist some even number greater than two that is not the sum of two primes.
CEP restriction clauses • To deal with these counterexample CEP can be equipped with these restriction clauses: • R 1: CEP can be used to derive ◇p from ◇cp only if p doesn’t contain any rigid designator, R, such that (i) R refers opaquely (i. e. , by describing an accidental property of its referent) and (ii) whether p is possible or impossible depends on R’s referent. • R 2: CEP can be used to derive ◇p from ◇cp only if p doesn’t make a mathematical or logical claim that, if false, could be refuted only by completing a supertask.
• But neither of these restriction clauses is of any help to Chalmers. • Cost G: acknowledging that the zombie argument’s proponent has a major unmet burden—namely, that of demonstrating the existence of a new category of non-Kripkean conceivable impossibilities, a category of which PT∧Q but not PT∧¬Q is a member.
The anti-CEP route • Let ◇sp represent the claim that p is ‘schmonceivable’, where this is a ‘catch all’ name for whichever epistemically accessible property of a proposition (perhaps primary ideal positive conceivability, but we leave this open) a proponent of the zombie argument who chooses the Non-Austere Route would opt to put in place of austere conceivability.
The schmonceivability argument against materialism
• Schmonceivability Entails Conceivability (SEC): ∀p (◇sp→◇cp) • Suppose SEC was false. Then there would be some proposition, p, that is schmonceivable but (austerely) inconceivable. But given p is inconceivable then, p is impossible. Hence p would be both schmonceivable and impossible, making it a counterexample to SEP, which the schmonceivabilist relies on to justify her own second premise, S 2. The schmonceivabilist can’t afford to admit SEP is undermined by some such counterexample. Hence she is committed to SEC’s being true
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