A Simple Question on the Consequence Argument
August 16, 2014 — 15:01

Author: Michael Almeida  Category: Uncategorized  Tags: , , , ,   Comments: 29

Here’s a fairly simple question about the consequence argument. Still, I think it is an interesting question that focuses on the conditionals in the argument. The rule α is not interestingly in doubt, but I assume both α and β. Let P* be  proposition describing a time slice of the universe at some point a billion years ago. Let L all the laws of nature. Let P be a true proposition describing any event after P*.  If determinism is true, then P follows from the conjunction of P* and L. ☐ stands for logical necessity. Assume that determinism is true. Here is van Inwagen’s argument.

1. ☐((P* & L) ⊃ P) assumption
2. ☐(P* ⊃ (L ⊃ P)) 1; modal, prop logic
3. N(P* ⊃ (L ⊃ P)) 2; rule β
4. NP* premise
5. N(L ⊃ P) 3, 4; rule α
6. NL premise
7. NP 5, 6; rule α

I’m not so concerned with working out whether rule α or β are valid. But I am concerned about some of the implications of the premises of the argument. Note first that (8) is necessarily true,

8. N(P* ⊃ (L ⊃ P)) ≣ N(P* & L) ⊃ P))

The conditional in (3) allows for strengthening antecedents. So, (9) is also a necessary truth, for any Q whatsoever.

9. N(P* & L) ⊃ P)) ⊃ N(P* & L & Q) ⊃ P))

So, (3), together with (8) and (9) entail (10).

10. N(P* & L & Q) ⊃ P))

Let Q be that God brings time to an end. Or let Q be that a miracle occurs, as certainly happens in some worlds. Or let Q be that an event happens between the occurrences of P* and P that is inconsistent with P, as certainly happens in some worlds.  Recall that Np is the proposition that “No one has any choice about the fact that p“. Now no one has a choice about time ending, or miracles occurring or events-inconsistent-with P occurring in the past, in addition to not having a choice about P* and L. And this is true whether or not any of these propositions in Q, L or P* are true.

Consider a world w in which Q is true. It is also true in w that (11), though P* and L are not both true there.

11. N(P* & L & Q)

There is no world in which I have a choice about P* or L, even if they’re false. So, we can derive that (12) is true in w, from (10) and (11).

12. NP & ~P

So, it is consistent with no one ever having a choice about, say, whether I raise my arm that I do not raise my arm. So, NP is consistent with my bringing about ~P. But then NP is consistent with a version of the principle of alternate possibilities that is relevant to my having free will relative to P.

• WH

Let Np be the proposition that “p, and no one has a choice about p”. Doesn’t this solve the problem you are referring to? I’m assuming you want to prove (11) by the agglomeration principle and N(P*) & N(L) & N(Q). The agglomeration principle is invalid, since Nx & Ny is consistent with ~N(x&y). Also, if you take Np as I have defined it, then N(Q) would seem to be false since Q seems to be false.

August 17, 2014 — 5:17
• Michael Almeida

Whether agglomeration is valid depends on whether your logic is normal or not. But that’s irrelevant here, since the inference to (11) is not the result of agglomeration; it results from strengthening antecedents. I don’t derive in the proof that NQ, I derive NP & ~P. It is true that we can alter the meaning of ‘Np’ to avoid the conclusion NP in worlds where L or P* are false. But, first, I have no choice about P, whether or not it’s true, so the change in meaning seems ad hoc. And second, the change does not preclude my having a choice about ~P* in worlds where P* is true, in particular, the actual world. But all I need is a choice about ~P* to have an alternative possibility (i.e., to be free in @).

August 17, 2014 — 8:21
• WH

I don’t see how you can prove (11) using strengthening antecedents. How can you prove N(P* & L & Q) using strengthening antecedents? I assumed you only think it’s true because [N(P*) & N(L) & N(Q)] imply it, and that since agglomeration is false, the implication is false (and then premise 11 would be unjustified). Can you sketch out a proof for (11) using your suggested method?

I’m not sure the change in meaning is ad hoc, because perhaps your definition of Np implies my definition given some reasonable assumptions. Let’s say that Np means “no one has a choice about p”. But having a choice about p is simply being able to do something that would make p false (when p is true of course).

In response to your second objection, it seems that to be free in any world, you must have dual control, so you must have a choice over P* and ~P*.

Note: your premise (3) follows from (2) and rule alpha (not beta).

August 18, 2014 — 9:59
• Michael Almeida

I do sketch a proof for (11). It follows from (3), (8), (9) and (10). Strengthening antecedents holds for strict conditionals, and N-conditionals follow from strict conditionals. It has nothing to do with agglomeration. For what it’s worth, agglomeration is not invalid tout court; again, it requires the assumption of a non-normal logic. I make no such assumption and need no such assumption. If we return to letting Np be the proposition that “No one has any choice about the fact that p“, then the proof yields (12), which was my aim. Whether freedom requires Kane-style dual control is beyond the scope of the discussion, which depends only on some version of PAP for freedom. I’m not interested in pursuing alternative necessary conditions for free action that never arise for van Inwagen.

August 18, 2014 — 10:16
• WH

3. N(P* ⊃ (L ⊃ P))
8. N(P* ⊃ (L ⊃ P)) ≣ N(P* & L) ⊃ P))
9. N(P* & L) ⊃ P)) ⊃ N(P* & L & Q) ⊃ P)) [which follows from strengthening the antecedent]

I can see how (3) & (8) & (9) entail (10).
10. N(P* & L & Q) ⊃ P))

But I can’t see how the conjunction of the above propositions, gets to (11).
11. N(P* & L & Q)

Anyway, (10) implies (12) without (11). All you need is (A): N(P*) & N(L) & N(Q).
(13) N[P*–>((L&Q)–>P)], from (10).
(14) N[(L&Q)–>P], from (13) and (A).
(15) N[L–>(Q–>P)], from (14).
(16) N(Q–>P), from (15) and (A).

(12) N(P) &~P, from (16) and (A), and since Q is true (in w) and Q–>~P.

So, I agree (12) follows without agglomeration. I was just confused about how (11) came about. But, again, (12) is only true if your definition of Np is true, because of my definition works, then (12) implies a contradiction. My definition is Np–>p, so N(P)–>P; combine this with (12) and you get P&~P.

I now remember where my definition of Np came from (not really “my” definition then). I read it in Alexander Pruss’ article “Incompatibilism Proved”. This is it:

“N-def. Nr if and only if r ∧ ∼ ∃x∃α[Can(x, α) ∧(Does(x, α) → ∼ r)]”

This is what he also had to say, “It is natural to read the claim that there was, is and will be no choice about r
as saying that there is nothing that anyone can (ever) do that would falsify r”.

As for dual control, it seems to me that dual control is mainly what makes incompatibilism attractive, and as such, it seems to be necessary for any incompatibilist account to include it. I accept that dual control is outside the scope of the discussion though.

One more thing, how does (12) imply a version of alternate possibilities relevant to having free will relative to P? N(P) & ~P is consistent with N(~P), isn’t it? Don’t you require ~N(~P) along with (12) to get your relevant alternative possibilities?

August 18, 2014 — 17:57
• Michael Almeida

3. N(P* ⊃ (L ⊃ P))
8. N(P* ⊃ (L ⊃ P)) ≣ N(P* & L) ⊃ P))
9. N(P* & L) ⊃ P)) ⊃ N(P* & L & Q) ⊃ P)) [which follows from strengthening the antecedent]

>>
(9) is necessarily true,following from the necessary truth ☐(P* & L) ⊃ P)) ⊃ ☐(P* & L & Q) ⊃ P)). The consequent differs from the antecedent in having a stronger antecedent.
<>
Right, (10) follows from those, as is evident in the proof.
<>
(11) follows from the meaning of ‘Np’ (as I have used it and defended it above)
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Hard to see how this is relevant. I do not assume (A) in the proof.

August 18, 2014 — 18:43
• WH

Right, I said that (9) DOES follow from strengthening the antecedent; I never denied that. However, I denied that (11) follows from strengthening the antecedent. I also denied that (11) follows from (3)&(8)&(9)&(10). I guess we ran through a misunderstanding, because we seem to be on the same page with regards to this; we just differ on (11), the definition of Np, the importance of dual control, and the consistency of (12) and a version of alternative possibility relevant to free will.

Anyway, (11) is quite different from (A). I see that you realize the difference. But given that I proved you can get from (10) to (12) without (11)- using the more plausible (A)- I don’t understand why you think (A) is irrelevant.
11. N(P* & L & Q).
(A): N(P*) & N(L) & N(Q).

Your Np simply states that “no one has a choice about p”. (11) follows from your Np only if no one has a choice about (P* & L & Q). But (A) says no one has a choice about P* and no one has a choice about L and no one has a choice about Q. (A) seems to capture the natural intuition regarding the consequence argument. After all, the initial consequence argument used N(P*) & N(L) and not N(P*&L). So (A) is more intuitively plausible than (11). I don’t really have a problem with (A); I’m just saying that your argument would be stronger if it includes (A) and not (11).

Also, your argument does not strictly include a defense of (11): “Now no one has a choice about time ending, or miracles occurring or events-inconsistent-with P occurring in the past, in addition to not having a choice about P* and L.” This can be translated as N(Q)&N(P*&L). This is neither (11) nor (A), but the point is that a natural reading of your argument may very well favor (A) over (11)- or, at least it doesn’t clearly favor (11) over (A).

August 19, 2014 — 8:40
• Michael Almeida

The argument is stated explicitly and does not include (A).

August 19, 2014 — 8:46
• WH

I know it doesn’t include (A). It includes (11) instead. I even retracted from my earlier claim that (A) entails (11). What I said is that it should include (A)- and not (11)- because (A) is more plausible than (11).

August 19, 2014 — 14:43
• WH

Correction: retracted my earlier claim…

Do you reject that (A) is more plausible than (11) and that your argument works just as fine without (11)?

August 19, 2014 — 14:48
• Mike:

Minor point: you’ve reversed rules alpha and beta. alpha is the rule that says that if necessarily p, then Np. beta is the transfer rule. In the following, I will stick to the terminology in the literature.

I’m with WH. Np is defined in the Consequence Argument literature as entailing p. Van Inwagen explicitly says: “Let ‘Np’ abbreviate ‘p and no one has, or ever had, any choice about whether p” (quoted in Finch and Warfield).

Let “N*” abbreviate your “no one has, or ever had, any choice about whether” operator. One can formulate rules analogous to alpha and beta for N*. The analogous rule alpha* will of course be entailed by alpha. But the analogous rule beta*, namely that if N*p and N*(if p then q), then N*q, does not seem to follow from beta. Indeed, beta* is clearly false. Let p be the claim that 2+2=5. Then N*p — no one has a choice about whether p. Moreover for any q, “if p, then q” is a necessary truth, and no one has a choice about it. So N*p and N*(if p, then q). But of course we don’t want to conclude that N*q.

Of course, van Inwagen denies beta now, in the light of what he considers to be definitive counterexamples. But consider beta2, which is Finch and Warfield’s plausible replacement for beta. beta2 says that if Np and necessarily(if p, then q), then Nq. We can make an analogous beta2*: if N*p and necessarily (if p, then q), then N&q. Again, beta2* is false: the same counterexample works as worked for beta*.

So, there is no plausibility in beta-like principles if you drop the p conjunct from the definition of Np.

August 21, 2014 — 9:51
• Michael Almeida

Thanks Alex. There’s good reason to drop beta, though it isn’t my concern here. My definition of ‘Np’ follows Mike Huemer’s in ‘Van Inwagen’s Consequence Argument’, Phil Review (2000), (Huemer reads ‘Np’ as no one has any choice about the fact that p. He takes that to entail p, but I’m less sure it does) but, as I noted above, I’m happy to adopt the version you propose. As I say there, it is true that we can alter the meaning of ‘Np’ to avoid the conclusion NP in worlds where L or P* are false. But, first, I have no choice about P, whether or not it’s true, so the change in meaning seems ad hoc. This seems to be a very important point: why would it turn out not to be the case that I have no choice about p in worlds where p is false? It just seems factually wrong. But second, if we use the conjunctive sense of ‘Np’, then the change does not preclude my having a choice about ~P* in worlds where P* is true, in particular, the actual world. But all I need is a choice about ~P* to have an alternative possibility (i.e., to be free in @). So, the assumption of the conjunctive ‘Np’ undermines the consequence argument.

August 21, 2014 — 10:03
• Michael Almeida

But the analogous rule beta*, namely that if N*p and N*(if p then q), then N*q, does not seem to follow from beta. Indeed, beta* is clearly false. Let p be the claim that 2+2=5. Then N*p — no one has a choice about whether p. Moreover for any q, “if p, then q” is a necessary truth, and no one has a choice about it. So N*p and N*(if p, then q). But of course we don’t want to conclude that N*q.

That’s a nice counterexample: but what’s it a counterexample to?. You certainly don’t want to conclude that it is false that no one has any choice about 2+2 = 5. No one has ever had a choice about the necessary truths or the necessary falsehoods, we can agree. So, what we have a counterexample to is the transfer of ‘no one has a choice about’. Whereas, if we adopt the conjunctive meaning of ‘Np’, it turns out false that no one has a choice about necessary falsehoods. But that’s difficult to believe.

So, here’s a counterexample for you. Let q = 2 + 2 = 5 and let p = any metaphysical impossibility. No doubt, p entails q, for any p. But since it is true that ~Nq (on the conjunctive reading of ‘N’), I can derive that ~Np, for all such p. But certainly for any metaphysical impossibility p, it is true that Np.

August 21, 2014 — 10:41
• Mike:

Huemer’s “no one has any choice about the fact that p” has a natural Russellian reading that entails that there is a fact that p. But that there is a fact that p entails

“Whereas, if we adopt the conjunctive meaning of ‘Np’, it turns out false that no one has a choice about necessary falsehoods”

I don’t see this. If we adopt the conjunctive meaning of “Np”, then for any necessary falsehood p we have ~Np. But ~Np doesn’t mean that someone has a choice about p. It means that either p is false OR someone has a choice about it.

August 22, 2014 — 10:20
• That said, I think there is a rather interesting thing in the vicinity that you’ve pointed out to me. Np = (p and N*p). But the intuition that beta (or beta 2) is true may be driven by intuitions about N*p rather than about Np. After all, N*p corresponds to a bit of ordinary language–“No one has a choice about whether p”–while Np does not (except on the somewhat forced Russellian reading). Thus, the intuition that beta (or beta 2) is true may be driven by the intuition that beta* (or beta 2*) is true. And the latter intuition is just plain wrong, as we all agree.

August 22, 2014 — 10:24
• Michael Almeida

I don’t see this. If we adopt the conjunctive meaning of “Np”, then for any necessary falsehood p we have ~Np. But ~Np doesn’t mean that someone has a choice about p. It means that either p is false OR someone has a choice about it.

So, you want to say that no one has a choice about metaphysical impossibilities but also that it is false that N(metaphysical impossibilities). Suppose that’s what you have in mind. In that case, no one has a choice about past truths or falsehoods, right? That’s what I’ve been pressing all along. In this case, adding the conjunct to the meaning of ‘Np’ insisting on the truth of p is ad hoc, it has absolutely nothing to do with the states of affairs one has no choice about. It plays no role in the consequence argument except to avoid some straightforward counterexamples.

August 22, 2014 — 10:34
• Mike:

The ‘Np’ operator is an explicitly stipulated operator. 🙂

We could work with N*p instead in the consequence argument. But then beta-2 would have to read:
If p and N*p and Necessarily(if p, then q), then N*q.
And beta, if we still want it, would have to be something like:
If p and q and N*p and N*(if p, then q), then N*q.

And then we avoid the problems you raise.

Then we are avoiding the counterexamples by having the conjunct p in beta-2’s antecedent. I think

August 22, 2014 — 11:37
• Mike:

The ‘Np’ operator is an explicitly stipulated operator. 🙂

We could work with N*p instead in the consequence argument. But then beta-2 would have to read:
If p and N*p and Necessarily(if p, then q), then N*q.
And beta, if we still want it, would have to be something like:
If p and q and N*p and N*(if p, then q), then N*q.

And then we avoid the problems you raise.

Then we are avoiding the counterexamples by having the conjunct p in beta-2’s antecedent. I think you might reasonably say that it is ad hoc to include this conjunct to avoid the counterexamples. I agree (see a post from today on my blog) IF the consequence argument is based on the sheer intuition that beta-2 is true. However, one can also run the consequence argument not on the basis of intuition, but on the basis of explicit proof from a counterfactual definition of “no one has, or has ever had, a choice about whether p” and axioms about counterfactuals. And that proof only establishes beta-2 with the conjunct p in the antecedent (as far as I know).

August 22, 2014 — 11:41
• Michael Almeida

The ‘Np’ operator is an explicitly stipulated operator
Alex,

I think my criticism of ‘Np’—i.e., that it is ad hoc—might depend on it being non-stipulative in the following sense: the operator aims to capture the unchoosability of certain propositions/state of affairs that (typically) have the ‘necessity of the past’. If that operator does not aim to capture that notion, then I don’t understand it. It is not as though the consequence argument is just an exercise in a formal system that has nothing to do with the actual metaphysics of the past. But since we are concerned with the metaphysics of the past–or the metaphysics of past states of affairs–we ought to employ the relevant sort of unchoosability operator. I think the right operator is N*p: it yields the right set of unchooseable states of affairs, Np does not.

August 22, 2014 — 16:14
• Mike:

…” the operator aims to capture the unchoosability”…

I don’t think so. One conjunct of the operator aims to capture unchoosability. The operator as a whole is just a convenient unit for the purposes of the argument.

August 24, 2014 — 21:32
• Michael Almeida

Alex,

The argument is not a logical exercise. It’s supposed to have important metaphysical implications, e.g. for freedom, responsibility, blame, etc. If the relevant operator does not exactly capture the actual necessity of the past (or, it’s fixity, or unchoosability, etc.), then the argument probably won’t establish anything important metaphysically.

August 25, 2014 — 9:56
• WH

What if N(L) means the laws are fixed? This is certainly not ad hoc. Also, it seems to me that the fixity of the laws implies that the laws exist and there is nothing we could have done that would have made them not exist. So, N(L) implies that the laws exist and N*(L), where this isn’t ad hoc.

Now, the defender of the consequence argument definitely wants to hold the following proposition:

(1) If the laws are fixed and determinism is true, then we don’t have free will.

If the fixity of the laws implies that (or is equivalent to) the laws exist and there is nothing we could have done that would have made them not exist, then we can accept (1) rationally.

But if the fixity of the laws is consistent with the laws not existing, and fixity of laws only means that we don’t control whether or not the laws hold, then we can’t accept (1) rationally. So, if fixity of laws only implies N*(L), then we can’t accept (1) rationally.

For, imagine that the laws ceased to exist at a certain time; perhaps God caused this to happen (God supposedly isn’t bound by the laws anyway even if N*p is true). It is still true that we don’t control whether or not the laws hold; only God does.
By the latter definition of “fixity”, the laws are fixed. Also, determinism holds since it is still true that if those laws are true, then we do a certain action. It seems consistent to say that because of the cessation of the laws, humans acquired libertarian free will (of course, this is true because we can consistently stipulate that there is no compulsion and reason-responsiveness holds, etc.). This story is definitely consistent with having libertarian free will even if acquiring libertarian free will is out of our control.

August 23, 2014 — 8:40
• Michael Almeida

What if N(L) means the laws are fixed? This is certainly not ad hoc.

It is not ad hoc if it is also true that N(metaphysical impossibilities), since those are also fixed.

(1) If the laws are fixed and determinism is true, then we don’t have free will.

So long as it is also true that past events are fixed (or at least the hard facts are fixed), then I agree they would believe (1).

But if the fixity of the laws is consistent with the laws not existing, and fixity of laws only means that we don’t control whether or not the laws hold, then we can’t accept (1) rationally.

I don’t think incompatibilists believe this. The consequence argument does not depend on the laws being necessary truths. What the fixity of the laws is understood to entail (post Lewis) is that the laws are counterfactually independent of our actions (i.e., we cannot violate a law and we cannot cause a violation of a law, but we also cannot do anything x such that, were we to do x, there would have been an exception to an (actual) law). So, among the things that are kept fixed in entertaining counterfactual worlds in which I, say, raise my hand at t, are worlds in which past facts are the same and the laws are unchanged. Since those worlds are impossible, I can’t raise my hand. This is not to say that there are no distant worlds in which I raise my hand and the laws are different. This is also not to say that it’s false that were I to raise my hand at t1 and a small miracle had occurred at t0, then the actual past would be different from what it is. It is to say that were I to raise my hand, it is false that a small miracle would have happened. That’s more or less the line the incompatibilists (or, to be on the safe side, many incompatibilists) take. In this sense, I have no choice about the laws, and so N*(L).

August 23, 2014 — 9:44
• WH

“It is not ad hoc if it is also true that N(metaphysical impossibilities), since those are also fixed.”

N(metaphysical impossibilities exist) is not fixed; whereas this is fixed: N(metaphysical impossibilities don’t exist).

“the fixity of the laws is consistent with the laws not existing.

>I don’t think incompatibilists believe this.”

I agree that they don’t believe this because it’s wrong. But this was my point: the fixity of the laws is not consistent with the laws not existing, so if the laws are fixed, then the laws exist (and there is nothing we could have done such that had we done it, the (actual) laws would not exist).

“The consequence argument does not depend on the laws being necessary truths”

I don’t see how you inferred that this is what I implied.

Let me make my main point again, without referring to fixity.

You want to affirm the following proposition:

(1*) If N*(L) and determinism holds, then we don’t have free will.

But N*(L) is consistent with ~L. So this implies:

(2*) If N*L & ~L & determinism, then we don’t have free will.

But (2*) is false, so (1*) is false.

To see that (2*) is false, I’ll re-post my story:

Imagine that the laws ceased to exist at a certain time; perhaps God caused this to happen (God supposedly isn’t bound by the laws anyway even if N*p is true). It is still true that we don’t control whether or not the laws hold; only God does. [So, N*(L) & ~L].

Also, determinism holds since it is still true that if those laws are true, then we do a certain action. [So, determinism holds].

It seems consistent to say that because of the cessation of the laws, humans acquired libertarian free will (of course, this is true because we can consistently stipulate that there is no compulsion and reason-responsiveness holds, etc.). This story is definitely consistent with having libertarian free will even if acquiring libertarian free will is out of our control. [Libertarian free will holds].

So, it is possible to describe a scenario where N*(L) & ~L & determinism & libertarian free will exists. So (2*) is false.

Notice, however, if N*(L) and L and determinism, then you can’t consistently say that libertarian free will exists in this scenario.

August 23, 2014 — 12:05
• Michael Almeida

N(metaphysical impossibilities exist) is not fixed; whereas this is fixed: N(metaphysical impossibilities don’t exist).

Ok, I’m now asserting the view I’ve been defending. In order to avoid the charge of being ad hoc, it would have to be true that N(metaphysical impossibilities) (where N is read ‘ is fixed’) since I can do nothing to make it true that 2 + 2 = 5.

You want to affirm the following proposition:

(1*) If N*(L) and determinism holds, then we don’t have free will.

No. I don’t want to affirm that. I don’t even want to affirm that If N(L) and determinism holds, then we don’t have free will.

August 23, 2014 — 12:50
• Michael Almeida

Imagine that the laws ceased to exist at a certain time…

For what it’s worth, the laws cannot cease to exist. The laws exist necessarily, if at all. Maybe you want to say that, possibly, the laws cease to be true. There’s no world w in which the laws L are true at t and cease being true at t+, but you can make sense (I think) of the laws of w ‘ceasing to be true’ in the sense that there is a world w’ which has the same pattern of law-like behavior as w up to t’, but diverges from w at t’.

August 23, 2014 — 13:06
• WH

“In order to avoid the charge of being ad hoc, it would have to be true that N(metaphysical impossibilities) (where N is read ‘ is fixed’) since I can do nothing to make it true that 2 + 2 = 5.”

N(p) only applies when p is a proposition. “Metaphysical impossibilities” is not a proposition. There are two propositions I assume you are referring to: either “metaphysical impossibilities don’t exist” or “metaphysical impossibilities exist”.

So either (a) N(metaphysical impossibilities exist) or (b) N(metaphysical impossibilities don’t exist).

My definition of Np: Np iff p and there is nothing anyone can do such that had he done it, p would have been false.
[where N is read as ‘is fixed’].

Whereas we take the definition (D1) of N*p as: N*p iff there is nothing anyone can do such that had he done it, p would have been false.

Or, we can take (D2) of N*p as: N*p iff either (p and there is nothing anyone can do such that had he done it, p would have been false) or (~p and there is nothing anyone can do such that had he done it, p would have been true).

Let’s take (a): N(metaphysical impossibilities exist).
This implies that metaphysical impossibilities exist. But metaphysical impossibilities don’t exist. Therefore, ~N(metaphysical impossibilities exist), therefore the existence of metaphysical impossibilities is not fixed. Yes, I stick to this argument, and I don’t see what the problem is with accepting that the existence of metaphysical impossibilities is not fixed.

Let’s take your example and apply it to (a):

N(‘2+2=5’ is true) implies that ‘2+2=5’ really is true, but it isn’t; therefore ~N(‘2+2=5’ is true). So (‘2+2=5’ is true) isn’t fixed. I definitely accept that.

Maybe, you might want to say: N*(‘2+2=5’ is true) because there is nothing we could have done that would have made ‘2+2=5’ false- on definition (D1) of N*p. This sounds wrong to me.

However, on (D2) of N*p, you get: N*(‘2+2=5’ is true) because there is nothing we could have done that would have made ‘2+2=5’ true. This is correct. But, I still affirm that (‘2+2=5’ is true) is NOT fixed (precisely because it is false). This means that we cannot read N* as ‘is fixed’; only N can consistently be read as ‘is fixed’.

Let’s take (b): N(metaphysical impossibilities don’t exist).
This implies that metaphysical impossibilities don’t exist and that there is nothing that we can do such that had we done it, ‘metaphysical impossibilities don’t exist’ would be false. Given that Np is read as ‘p is fixed’, then ‘metaphysical impossibilities don’t exist’ is fixed. Yes, all of this is true and coherent.

Let’s take your example and apply it to (b)- for completeness:

N(‘2+2=5’ is false) implies that ‘2+2=5’ is false and there is nothing we could have done that would have made ‘2+2=5’. Also, (‘2+2=5’ is false) is fixed. I accept all of this and believe it to be coherent.

On (D1) and (D2) of N*p, N*(‘2+2=5’ is false) because there is nothing we could have done that would have made ‘2+2=5’ true. This is correct. No problem here.

Note: my counterexample (the scenario with the laws ceasing to exist) applies to (D2) of N*p as well as (D1) of N*p.

August 23, 2014 — 16:10
• Michael Almeida

‘metaphysical impossibilties’ is a propositional variable in N(metaphysical impossibilities).

August 23, 2014 — 16:13
• Michael Almeida

the scenario with the laws ceasing to exist…

Once again, laws cannot cease to exist. It’s impossible. The thread is really losing value. I’m going to have to moderate soon.

August 23, 2014 — 16:21