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I believe the question is answered by the following paper; "A Note on Modal Formulae and Relational Properties", J. F. A. K. van Benthem, The Journal of Symbolic Logic, Vol. 40, No. 1 (Mar., 1975), pp. 55-58. DOI: 10.2307/2272270 URL: http://www.jstor.org/stable/2272270 which states: Theorem 1. There is no first-order formula $\phi$ such that $F ...


See : Alexander Chagrov & Michael Zakharyaschev, Modal Logic (1997), page 82 : A transitive frame $\mathfrak F$ validates the McKinsey formula iff satisfies the McKinsey condition where the McKinsey condition is : $\forall x \exists y(xRy \land \forall z(yRz \to y=z))$.


For a more basic coverage on the subject though, you can address to: Theodore Sider, Logic for Philosophy, Chap. 6, p. 178 There he provides simple proofs of the soundness, completeness and deduction theorem for a couple of modal systems he calls S. He does not tackle strong soundness and strong completeness though, since the notion of proof from a set ...


You can see : Raul Hakli & Sara Negri, Does the deduction theorem fail for modal logic (2010), for a detailed discussion of the Deduction Th in modal logic. See in particular page 6 for a discussion about an : argument for the failure of the deduction theorem [...] based on Kripke semantics. The "issue" with the proof of the Deduction Th for ...

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