domenica 26 gennaio 2020

definitive reflections

Here below I resume my definitive reflections on what firmly seems to me the complete and exhaustive solution of Goodman’s riddle.
(i)   something is grue iff (it is examined before t° and is green) or (it
is examined after t° and is blue)
(superfluous brackets are intended to facilitate  the argument  below).
BASIC POINTS:  Inclusive disjunction and range of quantification.
THEOREM. Claiming that *being green if examined before t°* ought to entail *being blue if examined after t°* is inconsistent.
CONSISTENCY CONDITION. No evidence can support at the same time incompatibile inductive generalizations.
PROOF . By reductio ad absurdum. Exactly as (i) defines *grue*, something is grink iff (it is examined before t° and green) or (it is examined after t°and pink) defines *grink*, and so on for every chromatic nuance we like (*grellow*, *grack* , *grurple*…). Therefore, if we claim that the observation before t° of  some green emerald is an inductive evidence for the generalization that all emeralds are grue, we must also claim that the same observation is an inductive evidence for the generalization that all emeralds are grink and grellow and so on.
Consequently if we claim that being grue entails being blue after t°, we must also claim that being grink entails being pink after t°, that being grellow entails being yellow after t° and so on. But as *being blue after t°*, *being pink after t°, *being yellow after t° * and so on are all incompatible predicates, we can conclude that our initial claim leads to inconsistency.
The consistency is restored  by a less superficial analysis of the matter. Here it is.
ANALYSIS. The definition of *grue* depends on four predicates, The adopted symbology is:
            R           grue
            G           green
            B           blue
            °           examined before t°
                       examined after t°
and then
                      green and examined before t° (I call it “gruebef”)
            B’          blue and examined after t° (I call it “grueaft”)
The couple of brackets occurring in (i) tells us that by definition the two predicates inclusively disjunct are not “green”  and “blue”  but “green and examined before t°” and “blue and examined after t°”. The various evidences we get by examining emeralds before t° are
(ii)  G°(a) & G°(b) & G°(c) & …
and then, by induction, the respective generalization is
(iii) (x) G°(x)
(all emeralds are gruebef).
Now, since a temporal opposition occurs in (i),  a basic question imposes itself: what is the range of the “all” (iii) speaks of? Is it the range of all emeralds examined before t° or  the range of all emeralds independently on the moment of their examination? Of course, as every evidence testified in (ii) refers necessarily to emeralds examined before t°, the inductive generalization cannot but refer to emeralds examined before t°; otherwise the definitory import of the temporal specification would be neglected. Our same intuition corroborates such a
position: the evidences we acquire by examining before t° the various emeralds cannot tell us anything about the color of emeralds examined after t° (“B” not even occurs in (ii)). Actually, owing to the inclusiveness of the disjunction occurring in (i), *being gruebef * is sufficient to derive *being grue*, that is
(iv)                    If (x) G°(x)      then (x)R(x)
(if all emeralds are gruebef, all emeralds are grue, and since all emeralds are gruebef, all emeralds are grue). But once more a question imposes itself: what is the range of “all”? An easy answwer, indeed: as the protasis of (iv) refers to all emeralds examined before t°, the apodosis too must refer to all emeralds examined before t°. The evidences we got do not allow any extrapolation to emeralds examined after t°. Thus the various definitions above (*grink*, *grellow*
etcetera) result perfectly compatible, and the riddle vanishes for, so to write, the ascertained gruebefness of emeralds does not allow us to expect any grueaftness (that is: does not allow us to state that emeralds examined after t° ought to be blue).
P.S. An apocryphal version of the riddle replaces the inclusive disjuncion with a conjunction. In this case, as actually being grue requires being blue after t°, the range of  quantification in All emeralds are grue should concerns emeralds examined both before or after t°. Yet being green before t° is no more a sufficient evidence for an emerald in order to conclude  its grueness, so that no riddle arises.

sabato 12 ottobre 2019


A premise. The occurrence of asterisks and inverted commas as diacritical symbols  is not  a typographical negligence,  but the consequence of well intentional reasons (here left out). Furthermore the inclusive disjunction, from Latin, is meant by “vel” and not by a potentially equivocal “or”.
The genesis of Goodman’s riddle can be outlined in three steps. First step. Blending two incompatibile current predicates (green/blue) in a just defined one (grue). Second step. Avoiding an open conflict between the two current predicates by a diachronical distinction (before/after a future moment t°). Third step. Violating the institutive assumptions by a mismanaged application of inductive generalization. The solution consists simply in restoring the coherence.
Now let me enter into details.
As for the definition of  *grue* I draw my track from Stanford Enciclopedia of Philosophy, voice “Nelson Goodman, §5.3. It runs as
(i)      something is grue iff (it is examined before t° and is green)
vel (it is examined after t° and is blue) (superfluous brackets for the sake of clearness).
THE RIDDLE. Owing to the above definition, before t° for each evidence statement asserting that a given emerald is green, there is a parallel statement asserting that the very emerald is grue. Thus the two inductive hypotheses *all emeralds are green* and *all emeralds are
grue* result equally confirmed. But as soon as we survey an emerald after t°, we knock against a contradiction, for the two hypotheses lead to opposite results.
And as far as I know, the worst aspect of the situation has not even been focalized. I mean that if we in (i) substitute “yellow” to “blue”
thus introducing by definition *grellow*, the same evidences above leading to *all eneralds are grue*, lead to *all emeralds are grellow* and so on for whatever color. The inductive inference from the observation of some green emeralds to the conclusion that all emeralds are grue (or grellow …)  precipitates us into a  chaos, more than into a riddle. It is absolutely wrong. Fortunately a sound appeal to logic overcomes the impasse.
CORRECT  REASONING. An observation represents an inductive evidence if and only if it sets up a particular instance of the generalization under scrutiny. The predicate  of the first inductive generalization, as (i) establishes explicitly, is  not “green” (simbolically “G”), but “examined before t° and green” (simbolically “G°”). This remark is not a
pedantry: the temporal specification is essential, since by definition the meaning of  “grue” depends strictly on it. Therefore, symbolically,
             G°(a) & G°(b) & G°(c) & …. → G°(x)
  is all what the inductive generalizations allows us to state; all emeralds examined before t° are green, therefore all emeralds examined before t° are grue. The extrapolation of this inductively legitimate conclusion  until
(ii)                    all emeralds are grue
smuggles a piece of information (all emeralds examined after t° are grue, then blue) quite arbitrary for supported by no evidence. In other words, (ii) states an absolutely ungrounded enlargement of the range “all” refers to.
Thus the riddle vanishes. The lesson is not the incompatibility between induction and (I quote SEP) “weird predicates” but the respect of the weird connotations imposed by definition to those predicates.
If a super-skeptic reader still had some doubt about the soundness of the argument claiming that the temporal conditions cannot be omitted, he ought to realize what follows. Let us remove the temporal conditions from (i); then the definition becomes
  (iii)                      something is grue iff (it is green) vel (it is blue)
and the inductive generalization (ii) follows unobjectionably from the greenness of the performed observations. Yet we are thus facing an obviousness, not a riddle. In fact from (ii) and (iii) we can infer that all emeralds are green vel blue, but such a conclusion, on the ground of the truth table for inclusive disjunction, agrees perfectly with the greenness of the performed observations: a green emerald, obviously, is also a green-vel-blue emerald.
The solution applies a fortiori to an apocryphal version of Goodman’s riddle. My strong purpose of concision induces me to neglect some other perhaps interesting considerations. Anyhow I am  ready to debate the theme.