Recently, I have been (re)reading Fitelson and Hawthorne's extensive and thorough discussion of "The Wason task(s) and the paradox of confirmation", which revived a puzzling little thought that I have. Wason's selection task is widely known, I guess. You have four cards as follows:
Each card has a letter on one side and a number on the other. A hypothesis is at issue, i.e., H = "for any card, if there's a vowel on one side, then there's an even number on the other side". Your task is to say which of the cards you need to turn over in order to find out whether H is true or false. (I want to be very careful here with the wording - I'm closely following Johnson-Laird & Wason, 1977, via Humberstone, 1994). [Edit: By the way - I forgot to say! - people strongly tend to choose the E- and the 4-card. The point is if this makes logical sense.]
Two cases have to be clearly distinguished: either (i) the domain of H is restricted to the four cards initially presented, or (ii) a larger deck is involved, from which those four cards have been sampled. If (ii) is allowed for, then very interesting issues arise (most famously tackled by Oaksford and Chater, 1994). However, Wason had (i) in mind, and everyone seems to agree that in that case the correct answer is: turn over the E- and the 7-card. Why?
From Wason on, the standard answer has been more or less the same: because only by turning the E- and the 7-card you could possibly falsify H. (Wason's experimental idea was apparently inspired by his knowledge of Popper.) But the logical point seems stronger than that: if the domain is restricted, then H is logically equivalent to "there's an even number behind E and a consonant behind 7". Logically equivalent - period. Not just such that it would otherwise be falsified. The proof is easy, and it provides ultimate motivation for why the selection of E and 7 is indeed compelling (recall how the question is spelled out: which cards are needed to find out whether H is true or false). Yet I have consistently found variants of the "Popp