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I'm starting to read Lewis' theory of counterfactuals. In his 1973 book, he specifies on page 10-11:

"The left-hand counterfactuls make trounle for the theory that the counterfactual is a strict conditional [...] if Ψ is true at every accessible Φ1-world, but not-Ψ is true at every accessible (Φ1&Φ2)-world, then there must not be any accessible (Φ1&Φ2)-worlds [...] Then if the lower counterfactuals are true, it is not thanks to their consequents: if a strict conditional is vacuously true, then so is any other with the same antecedent.

What I do not understand is how from the two premisses we derive that "then there must not be any accessible (Φ1&Φ2)-worlds [...]" and why this makes the next conditionals true in a vacuous way.

Geoffrey Thomas
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PwNzDust
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  • Because, by assumption, Ψ and not-Ψ would both have to be true at every accessible (Φ1&Φ2)-world, which is inconsistent. It is not clear what "lower counterfactuals" are and how the two parts of the quote are related. – Conifold Jul 08 '21 at 16:50

1 Answers1

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Lewis invites us to consider some counterfactual conditionals:

  1. Φ1 □→ Ψ

  2. Φ1 ∧ Φ2 □→ ¬Ψ

  3. Φ1 ∧ Φ2 ∧ Φ3 □→ Ψ

He points out that in the way such conditionals are ordinarily used, all of these may consistently be true, which is one way of saying that counterfactual conditionals are typically non-monotonic. This creates difficulties for the position that counterfactuals are strict conditionals, i.e. that they can be understood as conditionals in which the consequent part holds in every accessible possible world in which the antecedent part holds.

This is because if 1 is true, then Ψ holds at all accessible Φ1 worlds. But if 2 is true, ¬Ψ holds at all accessible Φ1 ∧ Φ2 worlds. This would require that both Ψ and ¬Ψ hold in every accessible Φ1 ∧ Φ2 world. The only way this would be consistent is in the trivial circumstance in which there are no accessible worlds in which Φ1 ∧ Φ2 holds. This would make 2 and 3 true, but only vacuously, since they would hold for any Ψ whatever.

Bumble
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  • Thanks for the extensive answer. There is still a point I do not find clear: why is the fact that Ψ and ¬Ψ hold in every accessible Φ1 ∧ Φ2 world consistent only if there are no accessible worlds in which Φ1 ∧ Φ2 holds? – PwNzDust Jul 09 '21 at 07:57
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    Ψ and ¬Ψ together form a contradiction, and in a normal modal logic, there are no worlds where contradictions are true. So, if every accessible Φ1 ∧ Φ2 world is one where Ψ and ¬Ψ both hold, then there can be no such worlds. You might think of it as a modal version of the fact that "all unicorns are white" and "all unicorns are not white" are only consistent if there are no unicorns. – Bumble Jul 09 '21 at 08:38