Reading my previous post again I realised that in (at least) one case I wrote confusingly. I try to be more clear in this.
Robin, the thing that I was trying to convey is that you misidentified the flaw in the p-zombie proof. The proof doesn't hold water. But the reason is not assuming the conclusion.
It is simply the first two of Geoff's statements with variables assigned. You can substitute back and get to the original.
Well, let's look at that original formalization with the modern syntax and see where it leads us:
[latex]$P_1$: $\forall x . (T(X) \rightarrow S(X))$ \\
$P_2$: $\neg S(a_1)$[/latex]
And consider the set P = { P
1, P
2 }. Now, this set P corresponds with a definition of p-zombies.
A definition of logical consistency is: A set of logical sentences is consistent if and only if it has at least one model. (You can state this in several equivalent ways).
The semantics of predicate calculus is defined in terms of structures, where you have an universe U and interpretations for the predicate symbols and function symbols that occur in the statements that we are examining. The idea is that the structure tells us when each predicate is true, effectively turning them to atoms of propositional logic. We then evaluate the statements using the truth values given by the structure. If it happens that all statements are true in the structure, then it is a model of the set.
In the case of P, we can create a structure that is a model for it easily. The simplest one is:
[latex]$U = \{ a_1 \} \\
$T = \emptyset$ \\
$S = \emptyset$[/latex]
Here P
1 is satisfied because its only instance is T(a
1) -> S(a
1) and T(a
1) is false, and P
2 is satisfied because S(a
1) is false.
So, we have here a model for P which means that P is consistent. Note that the model we found does not really represent any feature of the real world. Here we examined P alone, with absolutely no assumptions about the context where we might want to place it. And before adding in the context, let's examine P's logical consequences.
A formula f is a logical consequence of P if and only if adding its negation to P results in an inconsistent set. In this case we examine the formula neg T(a_1). So, we will now examine the set:
[latex]$P_1$: $\forall x . (T(X) \rightarrow S(X))$ \\
$P_2$: $\neg S(a_1)$ \\
$C$: $\neg \neg T(a_1)$
[/latex]
Next let's examine where this leads us:
[latex]$P_1$: $\forall x . (T(X) \rightarrow S(X))$ \\
$P_2$: $\neg S(a_1)$ \\
$C$: $\neg \neg T(a_1)$ \\
$N_1$: $T(a_1)$ (from $C$) \\
$N_2$: $T(a_1) \rightarrow S(a_1)$ (from instantiating $P_1$) \\
$N_3$: $S(a_1)$ (modus pones from $N_1$ and $N_2)
[/latex]
Since both S(a
1) and its negation are entailed, the set is inconsistent and neg T(a
1) is a logical consequence of P.
To recap: when we examine only P and absolutely no other assumptions whatever, we notice that P is consistent and it entails neg T(a
1).
Note that neg T(a
1) is not itself in P. We didn't assume it, it followed from P.
Next, we can try to see what happens when we include context for it. Since we are using simplified P, we might stay simple and consider only the formula:
[latex]$M_1$: $\forall x . T(X)$[/latex]
And define the set M = { M
1 } so that we don't later have to worry about mixing statements and set of statements. The meaning for M in the zombie context is "everything is physical".
Now M is also consistent and it is trivial to construct models for it.
But if we try to put P and M together, we get an inconsistency. None of the models of P are models of M and vice versa.
It is simply the first two of Geoff's statements with variables assigned. You can substitute back and get to the original.
Well, it is
a formalization of those statements (and roughly the same that Geoff intended). However, as I pointed out before, it is flawed. It is flawed because "x has y" is a binary property, something has something. When you formalize this using a unary predicate, you essentially move one of the arguments into the predicate symbol itself. So, instead of having "has(zombie,property)" as a predicate you now have "zombie-has(property)". This means that you have built the existence of zombies directly into the formalization itself so it is no wonder that it leads into a contradiction when the assumptions of materialism are added.
No. You need to think about this. Would the definition of a p-zombie be consistent if consciousness was physical?
The
definition of a p-zombie is the set P (well, the flawed definition, that is). As we saw before, the set P is consistent. So the only possible answer to this is: "Yes". You get only the contradiction when you add in M. So, you could have equally well used the red color to ask the question: "Would the definition of materialism be consistent if p-zombies existed?" And the answer to this too would be: "Yes", since M is a consistent set.
There is a way to phrase the questions so that the answer will be: "No". That would be: "Is P satisfied in any of the models of M?" or "Is M satisfied in any of the models of P?".
When we are examining pure mathematical logic, we have no reason to prefer the set M over P or P over M. The situation is completely symmetric. Any reason for us to choose one has to come from outside logic.
In the particular case of p-zombies we can escape the inconsistency by using an altered definition P' (that was presented some pages back) that doesn't assume the existence of zombies.
If the logical consistency of the definition depends entirely on the truth or falsity of the conclusion then you have assumed your conclusion.
Consider this final example that is perhaps
THE standard example of valid reasoning:
(a) All men are mortal.
(b) Socrates is a man.
And the conclusion that we get is "Socrates is mortal". Have we now committed the fallacy of assuming our conclusion? By the reasoning you used above, we have. Because if we add the sentence:
(c) Nothing is mortal
to the set we get an inconsistency. This is exactly the same that happened with P and M. Adding M to P results in inconsistency. Adding (c) to (a) and (b) results in inconsistency. So, if one is assuming the conclusion, then the other is also. Using your expression, "the syllogism is consistent only if we assume that its conclusion is true".
So, to recap, the problem with the p-zombie "proof" is that it has a hidden assumption that there are p-zombies. Not that it assumes its conclusion. If we remove this assumption, the inconsistency vanishes. Also, then we can't conclude that consciousness is not physical.