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Let me start with a few basic definitions.

Let $X$ be any set. A relation $R\subseteq X\times X$ is:

  • Euclidean when $$\forall x,y,z\in X: (x,y)\in R\,\wedge\,(x,z)\in R\, \rightarrow\,(y,z)\in R$$

  • Shift-Reflexive when $$\forall x,y\in X: (x,y)\in R\,\rightarrow (y,y)\in R$$

Claim. If $R$ Euclidean, then $R$ is Shift-Reflexive.

Pseudo-proof. Consider arbitrary $a,b\in X$. Assume that $(a,b)\in R$. Now, $(a,b)\in R$ and $(a,b)\in R$ [I know that it might seem weird to write it in that way, but I am stressing the point where I am using the antecedent of the definition of an Euclidean relation], therefore $(b,b)\in R$ by the fact that $R$ is Euclidean.

Attempted diagnosis of the problem with the pseudo-proof. The schema $$\forall x_1,x_2...,x_n: \phi$$ commonly used in these definitions (including the definitions I have provided above) is syntactically ambiguous, for in my view it might abbreviate either:

[1] $\forall x_1:\forall x_2:\,...\forall x_n:\phi$

or:

[2] $\forall x_1:\forall x_2:x_2\neq x_1:\,...\forall x_n:x_n\neq x_{n-1}\wedge x_n\neq x_{n-2}\wedge\,...\,\wedge x_n\neq x_1:\phi$

Now, if [2] is the case, then the pseudo-proof above is not a proof at all for obvious reasons, while, if [1] is the case, variables assignments need not to differ at each quantification instance, and therefore I don't see how the pseudo-proof above could fail to be an actual proof.

Now, if everything so far is correct, which is the correct interpretation for the definition of an Euclidean relation among the two proposed above (one consistent with [1], and another one consistent with [2])?

This is also related to the fact$-$in normal modal logic$-$that the schema 5 $$\lozenge p\rightarrow \Box\lozenge p$$ (valid in any Euclidean modal frame) does not imply (as far as I can see) the following schema (sometime referred to as $\,\Box M$): $$\Box(\Box p\rightarrow p)$$ (valid in any Shift-Reflexive modal frame).

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1 Answer 1

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1st part of the question: the answer is definitely that [1] is correct.

2nd part of the question: it seems it does imply. See Exercise 4.38 of Chellas' Modal Logic, 1980. It asks to prove that U (which you call $\Box M$) is a theorem of any normal $K5$-system.

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