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From: Philip Johnson-Laird BA PhD Psychology (UCL), Stuart Professor of Psychology Emeritus at Princeton. (Author isn't a logician.) How We Reason (1st edn 2008). p. 44.

  For premises that have several models, a single model consistent with them—an example—establishes a conclusion about what is possible; but all the models must be examples to establish a conclusion about what is necessary. The opposite is the case for refutations: a single counterexample refutes a conclusion about what is necessary, whereas all models must be counterexamples to refute a conclusion about what is possible. And so we should draw a conclusion about what’s possible faster than a conclusion about what’s necessary, but we should draw a conclusion about what’s not necessary faster than a conclusion about what’s not possible. This prediction is crucial for a theory that takes possibilities as fundamental, and several experiments have

p. 45

corroborated it. One caveat, learned the hard way, is that it’s no use asking adults whether something is possible if it's [Author's bungle. This ought be "its".] necessity is obvious. They baulk at saying, “yes”, in these circumstances; and so inferences have to be hard enough that when an event is possible its necessity isn’t obvious too. Here is a typical trial from an experiment. The premises are about a game of one-on-one basketball in which two players take part:

[1.] If Allan is in [the game] then Betsy is in.
[2.] If Carla is in then David is out.

The participants were more accurate and faster to infer that Betsy could be in the game than to infer that she must be in the game. If you list the possible games compatible with the premises, you’ll discover that there are three games and that Betsy is in all of them: Allan versus Betsy, Betsy versus Carla, Betsy versus David.

I refer to the players by only their initials.

  1. Why are A vs. C and A vs. D impossible?

  2. How can you efficiently deduce the emboldened sentence, without applying 1 and 2 to each person separately in order (1st: Alan v. B, C, D. 2nd: B v. C, D. 3rd: C v. D)?

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2 Answers 2

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A vs C and A vs D are impossible because A is in the game and B is not, violating [1].

If B is not in the game, then A cannot be in the game either without violating [1]. But then the players must be C and D, which violates [2]. So B must be in the game. All combinations with B are easily seen to be possible (any combination with B will satisfy [1], and to violate [2] you would have to have both C and D).

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Here's a formal proof in Fitch:

enter image description here

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