sabato 12 ottobre 2019

GOODMAN’S RIDDLE


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
follows:
(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.