# Rép : Modal Logic

```Le 13-août-05, à 08:38, Eric Cavalcanti a écrit :

```
```Can you give a particular example of a sentence p such that
B(Bp->p) is true?
```

```

```
Take any proposition you can prove, for example the tautology (p -> p), or t. Now we talk classical logic ok? So p -> (q -> p) is itself a tautology (the a fortiori
```axiom, which BTW is criticized by the Relevance Logicians and by the
Quantum logicians! (Note that it is accepted by intuitionist logicians)

Note for Lee:
```
The use of the "a fortiori axiom" has made me possible to simplify a bit recently some of the "intractable" problems, I say this for Lee Corbin who seems to infer I was stuck on necessarily intractable problem, it could still be that only my incompetence has to be blamed!).
```End note for Lee (apology to Eric).  <I must go now>

So once you have prove t, all the propositions of the shape

<anyproposition> -> t

```
are easily deducible, by applying modus ponens and the a fortiori axioms.
```
In particular Bt -> t is justified.

```
And thus B(Bt -> t) is true. (or, by necessitation B(Bt -> t) follows).
```
So an example of such a p making B(Bp -> p) true, is p = t.

But any p you actually prove will do.

Note:
```
I recall if necessary that Lob formula, B(Bp -> p) -> Bp, is *the* main axiom of G, the basic logic of arithmetical self-reference. You could search Lob or Santa Klaus in the archive. See the proof in the SANE paper.
```

Bruno

http://iridia.ulb.ac.be/~marchal/

http://iridia.ulb.ac.be/~marchal/

```