| if | a card has a D on one side | then | it has a 7 on the other |
| if |
P
|
then |
Q
|
|
D
|
F
|
7
|
5
|
|
P
|
not P
|
Q
|
not Q
|
|
Card #1
|
Card #2
|
Card #3
|
Card #4
|
Logical answer:
Given the truth table for "if . . . then . . ."
| Q is true | Q is false | |
| P is true | "If P then Q" is true | "If P then Q" is false |
| P is false | "If P then Q" is true | "If P then Q" is true |
a proposition of the form If P then Q is falsified if and only if P is true and Q is false.
Since we are looking for falsifying instances -- which are cases of P and not-Q -- we need to check anything that is P (to see if it might also be not-Q), and anything that is not-Q (to see if it might also be P). Things that are not-P and things that are Q are irrelevant.
Therefore, the correct answer, in the Wason trial above, is:
"Card #1 and card #4" -- because this corresponds to the instance
of P (card #1) and the instance of not-Q (card #4).
(Of course, in real experiments there are many trials, and the order of the cards is varied).
The analysis of case 2, below, is exactly parallel. However, many more people get the right answer in case 2 than in case 1: something about the content of the problem, as opposed to its form, makes the difference.
| if | someone drinks beer | then | (s)he is 21 or older |
| if |
P
|
then |
Q
|
|
beer
|
diet coke
|
23 years old
|
19 years old
|
|
P
|
not P
|
Q
|
not Q
|
|
Card #1
|
Card #2
|
Card #3
|
Card #4
|