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Tom
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| Posted 11/12/2003 4:37 AM (GMT -6) |   | Laika: If it's Sunday (a), Tom will go to church (b). Tom is in church (b). Does it necessarily mean it's Sunday (a)? No
Tom: Sounds good to me. So back to the drawing board on the EQs. Gosh.
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Tom
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| Posted 11/12/2003 6:29 AM (GMT -6) | | quote:
First—
This is definitely true: A --> ~B ^ ~C (“If A, then not-B and not-C”)
But does it follow that: ~B ^ ~C --> A (“If not-B and not-C, then A”)
Again—
This is definitely true: B --> ~A ^ ~C (“If B, then not-A and not-C”)
But does it follow that: ~A ^ ~C --> B (“If not-A and not-C, then B”)
Again—
This is definitely true: C --> ~A ^ ~B (“If C, then not-A and not-B”)
But does it follow that: ~A ^ ~B --> C (“If not-A and not-B, then C”)
It doesn’t seem to be the case that a --> b implies b --> a. For “If it’s the case that the sun is up, it’s the case that Tom is awake,” when true does not imply “If it’s the case that Tom is awake, it’s the case that the sun is up.” If the sun is up it may then rightly be inferred that Tom is awake, but Tom’s being awake does not necessarily imply that the sun is up.
So, it doesn’t seem that we can argue that the second argument in each of our three examples in the quote above follows from the truth of the first argument. Thus:
Qp --> ~Zp ^ ~Z~p does not imply ~Z ^ ~Z~p --> Qp. That is, “If indeterminately that p, then not determinately that p and not determinately that not-p” (which IS true) does not imply “If not determinately that p and not determinately that not-p, then indeterminately that p.” We can argue FROM “indeterminately that p” TO “might p” and “might not p” but we can’t argue FROM “might p” and “might not p” TO “indeterminately that p’.” Interesting.
This doesn’t mean that ~Z ^ ~Z~p --> Qp is false! All it means is that it doesn’t follow logically from Qp --> ~Zp ^ ~Z~p. But it sure makes it hard to prove that “might and might not” = “indeterminacy.”
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Alan Rhoda
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| Posted 11/12/2003 1:59 PM (GMT -6) | | I'll jump in for a bit. Have to defend my disseration in about a month, so not much time to dwell on technical details.
Tom, I see you found some great quotes from Prior. I'm gonna have to get my hands on a copy of that book.
Tom: I will argue that Aristotle and Prior are WRONG in claiming that if a proposition is contingent so is its contradictory. ... The contradictory of Qp is not Q~p; it is ~Qp. And Qp cannot imply ~Qp. It DOES imply Q~p (as Prior notes) but that’s what indeterminacy IS!
Alan: Right on, Tom.
Tom: 2. /QpvZpvZ~p (Either Qp or Zp or Z~p; if there’s a better way to notate it, let’s have it). We could formalize the implications of this too:
2a. Zp --> ~Qpv~Z~p
2b. Z~p --> ~Qpv~Zp
2c. Qp --> ~Zpv~Z~p
Alan: I think the "/" is unnecessary. If you have the "v" for "or" then "either" is implied. Furthermore, while Prior may like stringing symbols end-to-end, I find it more readable to add spaces between terms and conjunctions, viz. Qp v Zp v Z~p. Finally, I think your implications should say "and" and not "or". Thus:
2a. Zp --> ~Qp ^ ~Z~p
2b. Z~p --> ~Qp ^ ~Zp
2c. Qp --> ~Zp ^ ~Z~p
The reason for the "and" is that each of the three states of affairs is the metaphysical contrary of the other two. So if one is true, then both of the other two must be false.
Tom: 8. Qpq = Qpvp (or Qpq --> QpvQp). (I didn’t see this. I take it that if it’s indeterminately the case that ‘p and q’ (that is, Qpq), then it would follow that Qp^Qq, “It is indeterminately the case that p AND indeterminately the case that q.” Two indeterminacies don’t equal an either/or, they equal a both/and. Right?
Alan: My thinking was that if Q(p ^ q) then (p and q) is indeterminate, which will be the case when either of them is indeterminate. Now I see that this is not quite right because an "and" statement can be indeterminate in two different ways: (a) both the conjuncts are indeterminate, or (b) one conjunct is indeterminate and the other is determinately true. Thus,
Qp ^ Qq --> Q(p ^ q)
but the converse is not true. Rather,
Q(p ^ q) = (Qp ^ Qq) v (Qp ^ Zq) v (Zp ^ Qq) = (Qp v Qq) ^ ~(Z~p v Z~q)
Tom: 9. (Alan’s 8) Q(pvp) = Qp^Qq (or Q/pq = Qp^Qq) (I didn’t see this earlier. It’s confusing. But I now think this is right. It’s just another way of falsifying both Zpq and Z~pq. You’re saying, either p or q is indeterminate and we don’t know which is which. Hmm. I guess that DOES = Qp^Qq!).
Alan: Actually Q(p v q) does not say that either p or q is indeterminate; rather, it says that it is indeterminate either that p or that q. The scope of the indeterminacy is different. Now, as with the previous theorem, I no longer think this is quire right, for there are two ways in which an "or" statement can be indeterminate: (a) both disjuncts are indeterminate, or (b) one disjunct is indeterminate and the other is determinately false. Thus
Qp ^ Qq --> Q(p v q)
but the converse does not hold. Rather,
Q(p v q) = (Qp ^ Qq) v (Qp ^ Z~q) v (Z~p ^ Qq) = (Qp v Qq) ^ ~(Zp v Zq)
Tom: I’m still not sure when Z needs to be used. ... I mean, couldn’t we just remove the ‘Z’ operator from all your theorems and get the same results?
Alan: I've considered this, but I don't think we can remove the determinacy operator, unless we can find another way to keep ~Zp and Z~p distinct.
quote:
MIGHT: ~Z~p --> Zp v Qp (If “not determinately the case that not-p (= a true ‘might’)”, THEN EITHER “determinately the case that p” OR “indeterminately the case that p.”
MIGHT NOT: ~Zp --> Qp v Z~p (If “not determinately the case that p (= a true “might not”)”, THEN EITHER “indeterminately the case that p” OR “determinately the case that not-p.”
You're saying that when we conjoin ~Z~p ^ ~Zp (“might” AND “might not”), all we get is Qp (“indeterminately the case that p”). You're arguing that:
~Z~p ^ ~Zp --> Qp
Can't happen I don't think. You HAVE to conjoin EVERYTHING on the right side of BOTH “if/then” equations above as well....
Alan: I think that ~Z~p ^ ~Zp --> Qp. Here's why: First, informally speaking, the implication is commonsensical--if something is not determinate one way or the other, then it has to be indeterminate. What other option is there? Second, formally speaking, you are correct, Tom, that we need to add both sides of both equations. When we do we get
~Zp ^ ~Z~p --> (Zp v Qp) ^ (Z~p v Qp) = (Zp ^ Z~p) v Qp
but the first term, i.e., (Zp ^ Z~p), is necessarily false, leaving Qp as the only remaining option. In fact, not only does ~Z~p ^ ~Zp --> Qp, but the converse is true as well. So
~Z~p ^ ~Zp = Qp
In your demonstration, Tom, you combined the terms on the right incorrectly to get
Zp v Qp v Z~p v Qp
Instead, the middle "or" should be an "and", and the terms on either side of it should be grouped.
God bless,
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GregBoyd
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| Posted 11/12/2003 9:46 PM (GMT -6) | | Of course its false that A->B = B->A. "All apples are fruits "doesn't mean "all fruits are apples." But I don't think our dispute over conjoined mights comes down to this. Rather....
On the square, Qp falsifies Z-p. And Q-p falsifes Zp. So if both Qp and Q-p are true, both Z-p and Zp are false. Hence I argue conjoined mights -- independently true mights -- falsify "woulds". Your point, Tom, holds only if Qp entails Zp and Q-p entails Z-p. For then the conjoined would entail contradictory truth. But the square doesn't require this, I don't believe.
If it does, I supply the cohibas.
If it doesn;t, its you. Deal?
Still, even if I'm right, it exposes the weakness of the square. Indeed, this precipitated the move to a triangle in the first place. For as you hammer home, there's no single proposition to express indeterminacy as there is "determinacy."
Having waded into this conversation mid stream, I need to chew more on the epistemological/ontological distinction you feel is so crucial. Keep talking...
To Laika, I'm a little acquainted with Lukasiewicz and the multi-valued logics that flowed from his work. I was excited about it for a bit, but then, about a year ago, decided denying bivalence was not necessary to affirm indeterminacy. Might propositions ARE either true or false.
Doc
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Tom
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| Posted 11/13/2003 12:45 AM (GMT -6) | | Thanks Alan! We need you on this!
quote:
Alan: I think the "/" is unnecessary. If you have the "v" for "or" then "either" is implied. Furthermore, while Prior may like stringing symbols end-to-end, I find it more readable to add spaces between terms and conjunctions, viz. Qp v Zp v Z~p. Finally, I think your implications should say "and" and not "or". Thus:
2a. Zp --> ~Qp ^ ~Z~p
2b. Z~p --> ~Qp ^ ~Zp
2c. Qp --> ~Zp ^ ~Z~p
Right. I switched the v to ^ the last time I posted ‘em on the first page of this thread. Plus I numbered them 3, 4, and 5.
quote:
Alan: My thinking was that if Q(p ^ q) then (p and q) is indeterminate, which will be the case when either of them is indeterminate. Now I see that this is not quite right because an "and" statement can be indeterminate in two different ways: (a) both the conjuncts are indeterminate, or (b) one conjunct is indeterminate and the other is determinately true…
Tell me if I’m being too picky, but whatever is true is determinately true. The determinacy and indeterminacy we’re talking about is an attribute of the states of affairs being described and not the truth values of the propositions describing them. So, if it's indeterminately the case that p, then Qp is 'determinately true' (if one wishes to describe the certainty of our knowledge of the truth value). So, your last sentence there, “…and the other is determinately true” might better be said, “…and the other is determinate,” period, whether it’s “determinately the case that…” or “determinately the case that not.”
quote:
Thus, Qp ^ Qq --> Q(p ^ q) but the converse is not true.
I gotta think on this one. Why wouldn’t the converse be true? Q(p ^ q) is true if both p and q are contingent. If either p or q are not contingent, then the argument would be false. So Q(p ^ q) doesn’t seem to be equivalent to any argument in which either p or q is determinate (either ‘determinately that’ or ‘determinately that not’).
quote:
Rather,
Q(p ^ q) = (Qp ^ Qq) v (Qp ^ Zq) v (Zp ^ Qq) = (Qp v Qq) ^ ~(Z~p v Z~q)
How could Q(p ^ q) be equivalent to (Qp ^ Zq), to take one of your possibilities? As I noted above, doesn’t Q(p ^ q) mean it is contingent that p and it is contingent that q? The contingency attributed by Q to (p ^ q) is attributed to both p and q as a single unit. Then how could this be equivalent to an argument in which either Zp or Zq obtain, or one in which Z~p or Z~q obtain?
quote:
Tom: 9. (Alan’s 8) Q(p v p) = Qp ^ Qq (or Q/pq = Qp ^ Qq) (I didn’t see this earlier. It’s confusing. But I now think this is right. It’s just another way of falsifying both Zpq and Z~pq. You’re saying, either p or q is indeterminate and we don’t know which is which. Hmm. I guess that DOES = Qp^Qq!).
Alan: Actually Q(p v q) does not say that either p or q is indeterminate; rather, it says that it is indeterminate either that p or that q.
Hmmm. “Either p or q is indeterminate” means, in my mind, one or both of p and q is indeterminate, which is exactly how you formalize it below. But when you say “It is indeterminate either that p or that q” I hear epistemological indeterminacy. I hear, “Neither p is indeterminately the case nor q is indeterminately the case, nor are both indeterminately the case; it is determinately the case that p and determinately the case that not-q or determinately the case that q and determinately the case that not-p.” This makes perfect sense of “It is indeterminate either that p or that q.” If Q is an ontological-indeterminacy operator, then Q(p v q) has to mean “Either or both of p and q is indeterminate” which you nicely formalize the options of below. I don’t want to conflate ontology and epistemology in our use of Q. As you formalized what Q(p v q) is equivalent to it seems that you’re sticking to the ontological also.
quote:
Now, as with the previous theorem, I no longer think this is quire right, for there are two ways in which an "or" statement can be indeterminate: (a) both disjuncts are indeterminate, or (b) one disjunct is indeterminate and the other is determinately false.
Right. Are we using “or” in the inclusive sense as “either or both” or in the exclusive sense as “either but not both”? There are different ways of noting them formally. Doesn’t ‘v’ usually mean “either or both”?
quote:
Thus
Qp ^ Qq --> Q(p v q)
Hmm. But Q(p v q) allows for the possibilty of p or q being determinate. How can this follow from Qp AND Qq?
quote:
Q(p v q) = (Qp ^ Qq) v (Qp ^ Z~q) v (Z~p ^ Qq) = (Qp v Qq) ^ ~(Zp v Zq)
Exactly!
quote:
Alan: I think that ~Z~p ^ ~Zp --> Qp. Here's why: First, informally speaking, the implication is commonsensical--if something is not determinate one way or the other, then it has to be indeterminate. What other option is there? Second, formally speaking, you are correct, Tom, that we need to add both sides of both equations. When we do we get
~Zp ^ ~Z~p --> (Zp v Qp) ^ (Z~p v Qp) = (Zp ^ Z~p) v Qp
but the first term, i.e., (Zp ^ Z~p), is necessarily false, leaving Qp as the only remaining option. In fact, not only does ~Z~p ^ ~Zp --> Qp, but the converse is true as well. So
~Z~p ^ ~Zp = Qp
In your demonstration, Tom, you combined the terms on the right incorrectly to get
Zp v Qp v Z~p v Qp
Instead, the middle "or" should be an "and", and the terms on either side of it should be grouped.
Right on my main man! I see this now. Plus one of our very first theorems, Qp v Zp v Z~p, demands that one of these be true. If two are false then third has to be true.
The Revolution is rolling forward.
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Alan Rhoda
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| Posted 11/13/2003 12:54 PM (GMT -6) | | Doc: Having waded into this conversation mid stream, I need to chew more on the epistemological/ontological distinction you feel is so crucial.
Alan: I don't know if my take on this is exactly the same as Tom's, but here's an example that might help.
There's an ongoing debate among physicists between those who think that quantum mechanics reveals a real (i.e., ontological) indeterminacy in nature (as in the Copenhagen interpretation), and those who think the indeterminacy is merely epistemic, only apparent but not real (so argue David Bohm and his followers).
Now apply this to the debate over open theism. Is the future really (partly) indeterminate? Or is it only apparently indeterminate but in reality completely settled? In other words, when we say "might" is this just an admission of ignorance about the future (epistemic indeterminacy) or is it an affirmation of knowledge about the future (ontological indeterminacy)?
Now I don't see Tom's point that the "might" of the square is necessarily epistemic and not ontological. I think it can be interpreted either way depending on context.
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Tom
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| Posted 11/14/2003 3:21 AM (GMT -6) | | Alan: I don't know if my take on this is exactly the same as Tom's, but here's an example that might help.
There's an ongoing debate among physicists between those who think that quantum mechanics reveals a real (i.e., ontological) indeterminacy in nature (as in the Copenhagen interpretation), and those who think the indeterminacy is merely epistemic, only apparent but not real (so argue David Bohm and his followers).
Tom: Exactly.
Alan: Now I don't see Tom's point that the "might" of the square is necessarily epistemic and not ontological. I think it can be interpreted either way depending on context.
Tom: Regardless of what view one takes on the ontological nature of indeterminacy, it remains the case that "might" on the square negates "will not" but says nothing about the truth values of "will" and "might not." One is surely true and one is surely fasle. But just knowing the truth of "might" can't tell you which is which. So "might" speaks directly to an ontological state of affairs in that it falsifies "will not," but it only speaks epistemologically with respect to "will" and "might not." That's what I mean by "might's" functioning epistemologically.
Alan: My thinking was that if Q(p ^ q) then (p and q) is indeterminate, which will be the case when either of them is indeterminate. Now I see that this is not quite right because an "and" statement can be indeterminate in two different ways: (a) both the conjuncts are indeterminate, or (b) one conjunct is indeterminate and the other is determinately true.
Tom: I didn’t notice this when I first read it, but I’m looking for a definition of “and” that’s not inclusive, that is, one where “p and q” mean, “either one or both of p and q” (your (b)). “Or” can function inclusively (“either or both of p and q”) or exclusively (“either but not both”). But I’m not seeing how “and” is used exclusively as your (b) argues (“either but not both”, which is what your “one conjunct is Q and the other is Z means. Where did you get that? I understood that “and” is a simple conjunct that conjoins both variables and attributes equally to each whatever the operator specifies. This might be behind my objection to your formalizing of “and”: Q(p ^ q) = (Qp ^ Qq) v (Qp ^ Zq) v (Zp ^ Qq) = (Qp v Qq) ^ ~(Z~p v Z~q). If understood conjunct attributing to both p AND q the indeterminacy of Q, then Q(p ^ q) = Qp ^ Qq and that’s it. It can’t possibly be equivalent to Qp ^ Zq or to Zp ^ Qq. It’s “or” that allows those equivalencies.
And here’s your formalizing the “or”: Q(p v q) = (Qp ^ Qq) v (Qp ^ Z~q) v (Z~p ^ Qq) = (Qp v Qq) ^ ~(Zp v Zq).
I said “Exactly!” to this, but I’m thinking there’s more. Q(p v q) is also equivalent to (Qp ^ Zq) and (Zp and Qq) as well as the same with Z~p and Z~q which you do mention. If one indeterminate variable is all that’s needed to render Q(p v q) true, then the other variable may be anything OTHER than Q, that is, it has to be determinate. But determinacy gives us two options, “determinately that…” or “determinately that not-….” The determinate variable could be either and it would still be equivalent since one variable is Q. You only give the “determinately that” options in your equivalencies.
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Alan Rhoda
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| Posted 11/14/2003 2:47 PM (GMT -6) | | Tom: Regardless of what view one takes on the ontological nature of indeterminacy, it remains the case that "might" on the square negates "will not" but says nothing about the truth values of "will" and "might not." One is surely true and one is surely fasle. But just knowing the truth of "might" can't tell you which is which. So "might" speaks directly to an ontological state of affairs in that it falsifies "will not," but it only speaks epistemologically with respect to "will" and "might not." That's what I mean by "might's" functioning epistemologically.
Alan: OK, now I think I see your point, Tom. So "might" on the square functions partly ontologically and partly epistemologically, with the result being that use of the square tends to conflate the two. Interesting.
Now, back to the issue of Q(p ^ q) and Q(p v q).
First, a couple preliminary definitions/assumptions. (1) In indeterminacy logic, every proposition affirms that something is either determinately the case, determinately not the case, or indeterminately the case. In other words, (p)(Qp v Zp v Z~p). Thus, (2) compound statements like Q(p ^ q) and Q(p v q) need to be evaluated on a field of possibilities as follows:
As is evident from the diagram, Z(p ^ q) is the case if and only if Zp and Zq. And Z~(p ^ q) is the case if and only if either Z~p or Z~q. Thus, Q(p ^ q) is true in all other cases. These cases must satisfy two constraints: (a) neither Z~p or Z~q is true--for then Z~(p ^ q) would be true--and (b) Zp and Zq are not both true--for then Z(p ^ q) would be true. In all other cases, Q(p ^ q) is true. What are these cases?
Zp and Qq
Qp and Zq
Qp and Qq
That is why I said that
Q(p ^ q) = (Qp ^ Qq) v (Qp ^ Zq) v (Zp ^ Qq)
We can rewrite this in terms of the two constraints as
Q(p ^ q) = ~(Z~p v Z~q) ^ ~(Zp ^ Zq)
or equivalently as
Q(p ^ q) = ~(Z~p v Z~q) ^ (Qp v Qq)
Note from the diagram that: ~(Zp ^ Zq) = (Qp v Qq)
Similar remarks hold for Q(p v q).
Cheers,
Alan
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Tom
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| Posted 11/14/2003 7:13 PM (GMT -6) | | Alan: OK, now I think I see your point, Tom. So "might" on the square functions partly ontologically and partly epistemologically, with the result being that use of the square tends to conflate the two. Interesting.
Tom: Yeah, but it's not JUST the square. EQs ("epistemological quantifiers"--or if they're equivalent to them--"existential quantifiers") on the triangle do the same thing. You have ~Zp, ~Z~p, and ~Qp on the triangle. These posit what is ontologically the case with regard to what they falsify, but only what is epistemolgically the case with regard to what they affirm. It's just that on the square you only have TWO EQs (~Zp and ~Z~p), where we MUST have three (add on ~Qp) if we have three states of affairs. Each state of affairs must have a falsifier such that the other too states of affairs are unknown.
Alan: First, a couple preliminary definitions/assumptions. (1) In indeterminacy logic, every proposition affirms that something is either determinately the case, determinately not the case, or indeterminately the case. In other words, (p)(Qp v Zp v Z~p).
Tom: I LOVE this revolution. Just talking about something called "indeterminacy logic" is so awesome! Make us famous Alan!
Alan: Thus, (2) compound statements like Q(p ^ q) and Q(p v q) need to be evaluated on a field of possibilities as follows:
Tom: I love it. I have a couple of those too but haven't posted them yet. I'd disagree on a couple of the truth-values you assigned, but maybe that's because I'm defining "and" differently. You're still conceding that "and" can function like a disjuction meaning "either or both" while I think it is ONLY a conjunct "both and" at all times. I'll get back in a bit with which ones I'm having trouble with.
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Tom
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| Posted 11/14/2003 8:23 PM (GMT -6) | | Alan-
Let’s just look at the truth tables for one argument to see if we’re using ^ the same way:
I take ^ to be a conjunction meaning “both and”. I’m running through the few books I have on logic plus all the websites I can find for somebody that argues “and” can be equivalent to the inclusive “or” (“either or both”). Can’t find any. As far as I can discover “and” is a simple conjunct, period.
With that in mind, tell me if these truth tables make sense.
Back to your last post:
Alan As is evident from the diagram, Z(p ^ q) is the case if and only if Zp and Zq.
Tom: Right. That sounds to me like you just said: Z(p ^ q) = Zp ^ Zq. Isn’t this what you just said?
Alan: And Z~(p ^ q) is the case if and only if either Z~p or Z~q.
Tom: Right. Because the conjunct ^ means “both and” (and not ‘one might be true and the other false’ as you earlier argued), ~(p ^ q), as you say, demands BOTH Z~p and Z~q. To falsify an “and” you only need to falsify one of the variables. That is “either or both” of them may be false. But to prove an “and” you have to show that both are true. So, if Z~(p ^ q) means both Z~p and Z~q must be the case, then Z~(p ^ q) = Z~p ^ Z~q, right? Isn’t that what you just said?
Alan: Thus, Q(p ^ q) is true in all other cases.
Tom: Huh? Q(p ^ q) is only the case if both Qp and Qq. There are lots of other cases that don’t fit this.
In your table, on the left side that charts the possibilities for (p ^ q), under Q(p ^ q). If this means both Qp and Qq, then there’s only one combination in the entire column that is true, the last one, Qp and Qq. Likewise, under (p ^ q), Z(p ^ q) is only true in one case, just as you say. Though you see that ^ means “both and” and so does not permit any other combination of truth values in the case of Z, you don’t follow this in the case of Q. But Q is just as real, as impacting, a state of affairs as is Z and Z~. Why would only one possible combination express the truth of varialbes combined by "and" when each varialbe is Z, but multiple combinations be allowed when the same variables are Z~ and Q?
Equally then under (p ^ q), I would argue that Z~(p ^ q) is true in only one case, that in which both Z~p and Z~q. That's what "and" means. Because “and” is a conjunct that demands that all the variables embraced in the conjuntion be the case, it follows that any other combination of truth values for the variables named falsifies the argument. Same for the third of the "^" column. In these three columns only one in each column can be true, that which attributes the meaning of the operator to BOTH the varialbes.
It’s entirely different on the (p v q) side, because “or” can mean “either but not both” or “either or both.” Your table shows that you’re working with the latter meaning.
I agree completely with you Z(q v q) column. But for Z~(p v q)—which I understand to require either “or” (because v determines the truth values here) both Z~p and Z~q. If just either (or both) of these is in the equation then it’s true. So in this column you recognize ONLY the Z~p -- Z~q as true. I’d list as true for this column the following: the 2nd, the 4th, the 5th (your only), the 6th, and the 8th. This is because each of these have “either or both” of Z~p and Z~q. The third under (p v q) is Q(p v q), the third column, which I take to mean “either or both Qp and Qq.” So long as one of the variables is Q the argument is true. That’s what “or” means. So the true combinations in this column are the 3rd and the 6th-9th because each of these has at least one Q-variable in it.
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Tom
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| Posted 11/14/2003 11:44 PM (GMT -6) | | Just to clarify how I see things…
“Indeterminacy logic” is the logic of the future. Since the indeterminacy we’re developing this logic for is ontological, the future is the only thing that can possibly be in view, for both past and present are by definition determinate, they cannot now be other than what they are.
Having said that, when indeterminacy logic officially debuts in some published form, we’ll have to make clear why all variables (e.g., p, q, or r), whether affirmed or denied (e.g., ~p, ~q, ~r), are prefixed with one of two operators, Z or Q, and why the variables themselves represent propositions that are considered tenseless. In indeterminacy logic there is no p or q, or ~q or ~q, or any other variable affirmed or denied, without either a Z or a Q in front of it indicating futurity. In traditional propositional calculus, one CAN post p, meaning []p (if the brackets were a square, used in modal logic as a necessity operator), which would be equivalent to Zp in indeterminacy logic, but we can't do that in indeterminacy logic.
Also, “Z~” in Z~p is not an operator. It’s a state of affairs to be sure, but the negation defines the variable, not the operator: It’s Z + ~p. So, though we have three metaphysically distinct states of affairs (Zp, Z~p, and Qp), we have only two operators in indeterminacy logic, an operator for determinacy (Z) and one for indeterminacy (Q).
OK, so there are no operationally undefined variables in indeterminacy logic. Thus, one of our theorems:
Zp v Z~p v Qp, in which case,
Zp --> ~Z~p ^ ~Qp
Z~p --> ~Zp ^ ~Qp
Qp --> ~Zp ^ ~Z~p
Now, what is behind my understanding of Alan’s truth tables, why I changed the values, is because I understand a) the three states of affairs to be mutually exclusive (as Prior says, the “divide the field between them”) and b) I understand the conjunct “and” to be unequivocal (viz., it always means “both/and”), and c) “or” can be either inclusive or exclusive (to fill out the tables you have to choose one or the other, since the values for each would be different).
So, as for ^ (“and”):
Z(p ^ q) attributes Z to BOTH p AND q. That is, it attributes “determinacy” to both p and q.
Z~(p ^ q) attributes Z to BOTH ~p AND ~q. That is, it attributes “determinacy” to both ~p and ~q.
Q(p ^ q) attributes Q to BOTH p AND q. That is, it attributes “indeterminacy” to both p and q. Q~(p ^ q) is by definition equivalent to Q(p ^ q).
While as for inclusive v (“or” = “either or both of…”):
Z(p v q) attributes Z to EITHER or BOTH p and q. That is, it attributes “determinacy” to either or both p and q.
Z~(p v q) attributes Z to EITHER or BOTH ~p and ~q. That is, it attributes “determinacy” to either or both ~p and ~q.
Q(p v q) attributes Q to EITHER or BOTH p and q. That is, it attributes “indeterminacy” to either or both p and q. Q(p v q) is by definition equivalent to Q~(q v q).
If someone were employing “v” in the exclusive sense (“or” = “either but NOT both of…”), then:
Z(p v q) attributes Z to EITHER but NOT BOTH p and q. That is, it attributes “determinacy” to either but not both p and q.
Z~(p v q) attributes Z to EITHER but NOT BOTH ~p and ~q. That is, it attributes “determinacy” to either but not both ~p and ~q.
Q(p v q) attributes Q to EITHER but NOT BOTH p and q. That is, it attributes “indeterminacy” to either but not both p and q. Q(p v q) is by definition equivalent to Q~(q v q).
The only difference between tables expressing the truth values of the above two “or” arguments would be a difference in the value of the arguments when BOTH variables shared the operator's meaning. In the inclusive “or” table this argument would be true while in the exclusive “or” table this argument would be false.
Personally I think we ought to have ways of distinguishing between the inclusive and exclusive uses of “or.” Most of the (admittedly few) logicians I’ve read distinguish between these.
I hope this explains my disagreement with Alan’s tables.
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Tom
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| Posted 11/15/2003 2:59 AM (GMT -6) | | I’m going NUTS!
Theorems:
1. Zp v Z~p v Qp
2. Zp --> ~Z~p ^ ~Qp
3. Z~p --> ~Zp ^ ~Qp
4. Qp --> ~Zp ^ ~Z~p
5. Qp = Q~p
6. ~Zp = Z~p v Qp
7. ~Z~p = Zp v Qp
8. ~Qp = Zp v Z~p
The above 8 are the same 8 on the final list of theorems on the first page of this thread. These all seem to be agreed upon.
9-11 on that same page are still being debated. They weren’t really complete either. So I’m listing all three conjunctions and arguing their equivalencies:
Conjuntions:
9. Z(p ^ q) = Zp ^ Zq
10. Z~(p ^ q) = Z~p ^ Z~q
11. Q(p ^ q) = Qp ^ Qq
Similarly the disjunctions:
Disjunctions:
12. Z(p v q) = (Zp ^ Zq) v (Zp ^ Z~q) v (Zp ^ Qq) v (Z~p ^ Zq) v (Qp ^ Zq)
13. Z~(p v q) = (Z~p ^ Z~q) v (Z~p ^ Zq) v (Z~p ^ Qq) v (Zp ^ Z~q) v (Qp ^ Z~q)
14. Q(p v q) = (Qp ^ Qq) v (Zp ^ Qq) v (Z~p ^ Qq) v (Qp ^ Zq) v (Qp ^ Z~q)
Conjunctions falsified (not sure about these; they would be the EQs of the conjunctions 9-11). Now, what if the conjunctions 9-11 are falsified? Wouldn’t the falsification of #9 above be equivalent to the conjunctions of either #10 OR #11? Since we only know that #9 is false, we have to admit either #10 or #11 could be the true conjunction. And wouldn’t the falsification of #10 be equivalent to the conjunctions of either #9 or #11? And wouldn’t the falsification of #11 be equivalent to the conjunctions of either #9 or #10? They’d look like this:
15. (Falsified #9 above): ~Z(p ^ q) = Z~(p ^ q) v Q(p ^ q)
16. (Falsified #10 above): ~Z~(p ^ q) = Z(p ^ q) v (Q(p ^ q)
17. (Falsified #11 above): ~Q(p ^ q) = Z(p ^ q) v Z~(p ^ q)
Disjunctions falsified (not sure about these; they would be the EQs of the disjunctions 12-14). I think the same would apply to what could be known merely on the basis of falsifying one of our disjunctions.
18. (Falsified #12 above)—
~Z(p v q) = All the options that follow from #13 and #14 above?
19. (Falsified #13 above)—
~Z~(p v q) = All the options that follow from #12 and #14 above?
20. (Falsified #14 above)—
~Q(p v q) = All the options that follow from #12 and #13 above?
21. Former #12 on page one of this thread, Q(p --> q) = p --> Qq still seems right to me.
We could similarly argue:
22. Z(p --> q) = Zp ^ Zq
23. Z~(p --> q) = Z~p ^ Z~q
24. Q(p --> q) = Qp ^ Qq
We have to think about how to related codependent operators:
25. Qp --> Zq
This is certainly NOT the way to say what I want. What I want to say is “p is indeterminately the case; but IF p, then q will necessarily follow”? So, BOTH p and q are Q, but q is Z IF p. This is interesting.
We need an EVENTUALITY OPERATOR! Hot dog! How do you describe what necessarily follows from what is itself indeterminately the case? Hmmm? Alan?
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| Posted 11/15/2003 6:06 AM (GMT -6) | | Hi all-
Here’s my version of Alan’s table:
Doc suggested that the notation excludes a lot of folks from discussion. Sorry ‘bout that all. In my obsessing I just forgot the whole world. I’d like to take the theorems we have so far and spell ‘em out so everyone can follow and those who want can participate. Sorry y’all.
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| Posted 11/16/2003 4:48 AM (GMT -6) | | Check out these two quotes from David Lewis’s Counterfactuals:
“I believe that there are possible worlds other than the one we happen to inhabit. If an argument is wanted, it is this. It is uncontroversially true that things might be otherwise than they are. I believe, and so do you, that things could have been different in countless ways. But what does this mean? Ordinary language permits the paraphrase: there are many ways things could have been besides the way they actually are. On the fact of it, this sentence is an existential quantification. It says that there exist many entities of a certain description, to wit ‘ways things could have been’. I believe that things could have been different in countless ways; I believe permissible paraphrases of what I believe; taking the paraphrase at its face value, I therefore believe in the existence of entities that might be called ‘ways things could have been’. I prefer to call them ‘possible worlds’.” (p. 84)
To which I’d like to say to Lewis: If there exist ‘ways things could have been’ then there must now exist ‘ways things might and might not yet be’.
Interesting. But as you read through Counterfactuals you notice that Lewis never thought of applying this to the future. He hints at it a couple of times (possibly in the quote below), but for the most part he talks about possibilities in terms of what “might have” or “would have” been. Amazing. The indeterminacy logic we’re trying to express now is all there in past tense in Lewis. We just need determinacy and indeterminacy operators (Z and Q!) to provide the means for discussing these ‘ways things might and might not yet be’.
Here’s one passage in Counterfactuals that seems to indicate that Lewis thought of the future in determinate and indeterminate terms. He even created what seems to be to be an indeterminacy operator like our Q. Check it out:
“Suppose that our language has the following resources: (1) propositional variables, grammatically interchangeable with sentences; (2) existential and universal quantifiers E [should be a backwards ‘E’ but I can’t type it] and A [should be an upside down ‘A’ but I can’t type that either] that may be used to bind these variables; (3) the operators [] [a square, but I can’t type it] and * [the * should be a diamond shape figure but I can’t type that either] of…necessity and possibility; (4) the truth-functional connectives; and (5) a special one-place sentential operator O [not the capital ‘O’ but a circle; I can’t type it], called the sphericality operator. A sentence Op is to be true at a world i if and only if there is some sphere S around i such that p is true at all and only the worlds in S.” (p. 45)
Looks to me like Lewis is talking about three operators: necessity (the operator for which is a small square), possibility (the operator for which is a small diamond), and what he calls sphericality (the operator for which is a small circle). If this were the traditional square, the necessity operator would run across the top as "will" and "will not" and the possibility operator would across the bottom as "might" and "might not." THere IS NOT PLACE on the sqaure for Lewis's sphericality (which sounds to me like our 'indeterminacy' operator.
Lewis talks about “spheres of accessibility” that I understand to be all the possible worlds accessible from this world i. Lewis’ O sounds like our Q, a future event that is neither necessary nor impossible. “It’s indeterminately the case that p” can only be true if p is accessible, actualizable, obtainable from THIS world. If you can get to p from where you are now, and if p is neither inevitable nor impossible, then Op (as Lewis notes it) or Qp (as we’ve been saying it) is true. If you can’t get there, then it’s not true. Not sure if I’m reading Lewis correctly, but given what he says in the first quote about believing this world might have turned out different in many ways, it makes sense to understand his Op as our Qp, as positing an event which might and might not happen.
Tom
PS I think, Alan, I'm getting a better grip on quantification. I earlier said Q states of affairs were not quantifiable, but now that I understand how “might” (the existential quantifier) and “might not” (the universal quantifier) work, it would seem that ~Qp IS a quantifier of Qp, just as ~Zp is the universal quantitifer of Zp and ~Z~p is the existential quantifier of Z~p. Not sure though.
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| Posted 11/16/2003 8:17 AM (GMT -6) | | A few more theorems for the nerds out there. These are something like “Theorems of Conjoined Quantifiers.” The quantifiers are the three (formerly termed by me) EQs. They are the “might,” the “might not” and the “not indeterminate that not,” the three props that negate one of the three states of affairs and say nothing certain about the truth values of the other two. If these pass (and I think they do) then we’ll add ‘em to our list.
1. ~Z~p ^ ~Zp = Qp
2. ~Z~p ^ ~Qp = Zp
3. ~Zp ^ ~Qp = Z~p
I think we already said (1) somewhere. It’s Doc’s “might and might not”
1. “Not determinately that not-p” and “Not determinately p” = “Indeterminately that p”
2. “Not determinately that not-p” and “Not indeterminately that p” = “determinately that p”
3. “Not determinately that p” and “Not indeterminately that p” = “determinately that not-p”
“Might” and “might not” are not the only two quantifiers that can be conjointly true to yield a true state of affairs. There are three such combinations. If you conjoin any two quantifiers, the state of affairs that lies directly in between them is the true state of affairs.
Cool.
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| Posted 11/16/2003 5:06 PM (GMT -6) | | I think this might help clarify a few things. Ha!
I was hoping this would help by listing all the different ways things are referred to.
The (1) of each point of the two triangles all relate together, similarly the (2)’s, (3)’s, etc. In modal logic the square and the diamond are already established operators for “It is necessary that…” and “It is possible that….” The Op is supposed to be a geometric circle and not the letter “O” but I couldn’t draw a circle. This “O” (circle) is David Lewis’s invention, and I think it parallels our indeterminacy operator “Q”. The (2) of each is the notation we’ve come up with here on our own. The (3) is a definition of (1) and (4) is a definition of our (2). (5) is the way normal folk say it, and (6) gives the epistemological implications.
Note: (3) on the lower left and lower right corners should say (on the left) “Possibly p” (not necessarily not-p)” and (on the right) “Possibly not-p (not necessarily p).” The “possibly” is for the possibility operator (the little diamond shaped buger). I forgot to include that.
For the sick souls still following any of this, it should be clear that the traditional square of counterfactuals, that looked like this:
“will p” --- “will not p”
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“might p” – “might not p”
Beside the fact that the William doesn’t have to pee at all, this could be rewritten with the little square (necessity operator) across the top, and with the little diamond shaped buger (possibility operator) across the bottom. If both possibility operators across the bottom are true, then you have (I think) what Lewis calls “sphericality” (our “indeterminacy”) with his circle as an operator. We’re using the letter Q for this.
You know, come to think of it, I think that in his head Lewis dumped the square too. He just did it quietly and didn’t announce it. Hmm.
Believe it or not, I think I’m out of gas. Nothing more to say of obsess with. It’s all basically out of my head and in these threads. I wanted a place to store it all and come back to it later. Now that’s done. I’ll let it rest a spell  . Sorry to put you all through it with me.
Have fun!
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| Posted 11/17/2003 3:01 AM (GMT -6) | | Alan-
Forgive me, but I’ve been thinking through your truth table for “and” and “or” again. I posted my own too, but I’m rethinking it. I still think you were earlier incorrect to argue that “and” isn’t limited to meaning “both and,” so I think that messed with you tables. But at the same time I think I misread the arguments too. Do me a favor. Assuming “and” means “both and” and never means “either or both,” would you say I’m understanding the following arguments correctly?
Z(p ^ q) attributes Z to “p and q” taken together. That is, “p and q” WILL both obtain. All that’s needed to falsify this is for either Zp or Zq to fail to appear in our argument.
Z~(p ^ q) attributes Z to “Not ‘p and q’ ” (‘p and q’ taken together that is). So, “p and q” together WILL NOT obtain. This doesn’t mean p cannot obtain alone or that q cannot obtain alone. It means that both p and q together will not obtain. All that’s needed to falsify this is for both Zp and Zq to obtain.
Q(p ^ q) attributes Q to “p and q.” That is, “p and q” taken together, as a whole, might obtain together and might not obtain together. So all that’s needed to falsify this is for either Zp and Zq to both obtain or for Z~p and Z~q to both obtain.
If I’m now understanding these correctly, then I’d change my truth tables, but I don’t think they’d agree completely with yours.
OK, now for the “or” arguments.
Z(p v q) attributes Z to “p or q” taken together, as a single entity. That is, “p or q” will obtain. All that’s needed to prove this is for either Zp or Zq, or both Zp and Zq, to obtain. Anything else falsifies it.
Z~(p v q) attributes Z to “Not ‘p or q’ ” (‘p or q’ taken together as a single entity). So, “p or q” together cannot obtain. All of the possible arrangements that would make (Zp or Zp) true would in the case of Z~(p v q) falsify it. Therefore, all that’s needed to falsify Z~(p v q) is for either Zp or Zq to obtain.
Q(p v q) attributes Q to “p and q.” That is, “p and q” taken together as a whole, might obtain and might not obtain. So all that’s needed to falsify this is for either Zp and Zq to both obtain or for Z~p and Z~q to both obtain.
I’m not having any problems understanding “and” or “or.” The question I have is how the meaning of our operators Z and Q and ~ when they stand outside the parenthesis of an argument, distribute their meaning over the variables inside the parenthesis. Maybe I’m not sure how you’re using parenthesis. Not everybody employs them with precisely the same meanings. If you can clarify your understanding of the above six arguments for me I’m sure we’ll agree on the truth tables.
When you get the time. I don't wanna interrupt your defense of your thesis. I'll live until you can get back to me...for a couple of days at least!
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| Posted 11/17/2003 3:11 AM (GMT -6) | | Laika-
Thank you Bro! Now we can turn out some nice looking anal-retentive lingo! Way to go. I was gonna ask the moderator if logical notation, like the ones you included and some others, could be added to the list of "Smilies" too the left of the editing window. If you want a necessity operator you just click it on and, presto, it's in your text between brackets. I wonder if that is doable.
I also noticed that I misplaced the universal and existential quantifiers on my last triangle. Rats. Never mind. It'll change before too long!
I'm cookoo for Coco-Logic.
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Tom
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| Posted 11/17/2003 4:09 AM (GMT -6) | | Laika: Tom, I didn't insult the prophet or anything like that, did I? 
Tom: You just sold me your house for $50.00.
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| Posted 11/17/2003 4:13 AM (GMT -6) | | | | |
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