Help with modal logic exercise

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I'm teaching myself modal logic, and I'm curious about the following exercise, taken from this book...

Homework Statement



Prove [tex]T:\; \Box A \rightarrow A[/tex] using the following rules:


Homework Equations



Given the structure [tex]M=<W,R,P>[/tex] in which [tex]W[/tex] and [tex]P[/tex] are as they are in a model, and [tex]R[/tex] is an equivalence relation on [tex]W[/tex], we have:

(1) [tex]\vDash_{\alpha}^M\mathbb{P}_n[/tex] iff [tex]\alpha\in P_n[/tex], for [tex]n=0,1,2,...[/tex]

(2) [tex]\vDash_{\alpha}^M\mathbb\top[/tex]

(3) Not [tex]\vDash_{\alpha}^M\bot[/tex]

(4) [tex]\vDash_{\alpha}^M\neg A[/tex] iff not [tex]\vDash_{\alpha}^MA[/tex]

(5) [tex]\vDash_{\alpha}^MA\wedge B[/tex] iff both [tex]\vDash_{\alpha}^MA[/tex] and [tex]\vDash_{\alpha}^MB[/tex].

(6) [tex]\vDash_{\alpha}^MA\vee B[/tex] iff either [tex]\vDash_{\alpha}^MA[/tex] or [tex]\vDash_{\alpha}^MB[/tex], or both.

(7) [tex]\vDash_{\alpha}^MA\rightarrow B[/tex] iff if [tex]\vDash_{\alpha}^MA[/tex] then [tex]\vDash_{\alpha}^MB[/tex].

(8) [tex]\vDash_{\alpha}^MA\leftrightarrow B[/tex] iff [tex]\vDash_{\alpha}^MA[/tex] if and only if [tex]\vDash_{\alpha}^MB[/tex]

(9') [tex]\vDash_{\alpha}^M\Box A[/tex] iff for every [tex]\beta\in M[/tex] such that [tex]\alpha R \beta[/tex], [tex]\vDash_{\alpha}^MA[/tex].

(10') [tex]\vDash_{\alpha}^M\Diamond A[/tex] iff for some [tex]\beta\in M[/tex] such that [tex]\alpha R \beta[/tex], [tex]\vDash_{\alpha}^MA[/tex].


The Attempt at a Solution



I'm unsure of the proper notation to begin this. I would think that we would begin by declaring a variable to represent the actual world. Let's use [tex]\alpha\in W[/tex]. Then for [tex]M=<W,R,P>[/tex] as described above:

[tex]\Box A\rightarrow\vDash_{\alpha}^M\Box A[/tex]

Correct?

And if this is the case, then by (9') and the reflexivity of [tex]R[/tex] (all equivalence relations are reflexive), we can say:

[tex]\vDash_{\alpha}^M\Box A[/tex] with [tex]\alpha\in M[/tex] and [tex]\alpha R \alpha \;\rightarrow\; \vDash_{\alpha}^MA[/tex]

And since [tex]\alpha[/tex] is the actual world, we have [tex]\vDash_{\alpha}^MA\rightarrow A[/tex]. Correct?

So, assuming all this is proper, then we give the following proof...

Choose [tex]\alpha\in W[/tex] with [tex]\alpha[/tex] the actual world and [tex]M=<W,R,P>[/tex] as described above. Then:

[tex]\Box A\rightarrow \vDash_{\alpha}^M\Box A[/tex] with [tex]\alpha\in M[/tex] and [tex]\alpha R \alpha \;\rightarrow\; \vDash_{\alpha}^MA\rightarrow A[/tex]


Does that look okay?
 

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