Help with modal logic exercise

[hatsoff]

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

Homework Statement

Prove $$T:\; \Box A \rightarrow A$$ using the following rules:

Homework Equations

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

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

(2) $$\vDash_{\alpha}^M\mathbb\top$$

(3) Not $$\vDash_{\alpha}^M\bot$$

(4) $$\vDash_{\alpha}^M\neg A$$ iff not $$\vDash_{\alpha}^MA$$

(5) $$\vDash_{\alpha}^MA\wedge B$$ iff both $$\vDash_{\alpha}^MA$$ and $$\vDash_{\alpha}^MB$$.

(6) $$\vDash_{\alpha}^MA\vee B$$ iff either $$\vDash_{\alpha}^MA$$ or $$\vDash_{\alpha}^MB$$, or both.

(7) $$\vDash_{\alpha}^MA\rightarrow B$$ iff if $$\vDash_{\alpha}^MA$$ then $$\vDash_{\alpha}^MB$$.

(8) $$\vDash_{\alpha}^MA\leftrightarrow B$$ iff $$\vDash_{\alpha}^MA$$ if and only if $$\vDash_{\alpha}^MB$$

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

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

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 $$\alpha\in W$$. Then for $$M=<W,R,P>$$ as described above:

$$\Box A\rightarrow\vDash_{\alpha}^M\Box A$$

Correct?

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

$$\vDash_{\alpha}^M\Box A$$ with $$\alpha\in M$$ and $$\alpha R \alpha \;\rightarrow\; \vDash_{\alpha}^MA$$

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

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

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

$$\Box A\rightarrow \vDash_{\alpha}^M\Box A$$ with $$\alpha\in M$$ and $$\alpha R \alpha \;\rightarrow\; \vDash_{\alpha}^MA\rightarrow A$$


Does that look okay?

 

[mighty2000]

it is great to see someone teaching themselves modal logic! It is a very interesting and useful branch of logic. I will provide a proof for the given exercise below:

Proof: Let \alpha be the actual world and M=<W,R,P> as described above.

Assume \vDash_{\alpha}^M\Box A. By (9'), for every \beta\in M such that \alpha R \beta, we have \vDash_{\alpha}^MA. Since R is an equivalence relation, \alpha R \alpha, which means \vDash_{\alpha}^MA.

By (7), if \vDash_{\alpha}^MA, then \vDash_{\alpha}^MA\rightarrow A. Therefore, by modus ponens, we have \vDash_{\alpha}^MA\rightarrow A.

Since we assumed \vDash_{\alpha}^M\Box A, and we have shown that \vDash_{\alpha}^MA\rightarrow A, we can conclude \vDash_{\alpha}^M\Box A \rightarrow A by (7).

This concludes the proof. Keep up the good work learning modal logic!