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Note:

There is a lack of consistency regarding symbols for modal operators. Hughes and Cresswell use a different symbolization schema than many and a comment about L & M may provide clarity for reader unfamiliar with their notation.

They use L for the necessity operator.
They use M for the possibility operator.

So for those not familiar with this scheme: L = ◻ and M = ◇

I just want to see if I got this correct.

Show that the following wff is valid in every seating arrangement: Mp -> (Lq->Mq).

Here's my argument:

Assume player A sees at least one player B that raises his hand for p, and thus also A raises his hand for Mp.

Assume player A doesn't raise his hand for Lq ->Mq, but he raises his hand for Lq but not for Mq.

Then player A sees all other players raising their hands for q, and thus A raises his hand for Mq, contradiction.

Another question from the same book.

exercise 1.2: show that in any seating arrangement in which there's a player who cannot see himself Lp->p isn't valid.

My answer: Suppose A cannot see himself. Let p be on every player's sheet except A's. Suppose that A raises his hand for Lp->p then he must raise his hand for p, since he raised his hand for Lp, but he cannot see everyone's sheet (only his own), so he should keep his hand down for p, and we get that Lp->p isn't valid.

Is my reasoning ok or not?

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  • It might be useful for people who are familiar with modal logic, but not the specific book you're using, to explain what the operators L & M stand for; I'm not sure that terminology is as standardised for modal logics as it is for say first-order logic.
    – Mozibur Ullah
    Commented May 20, 2015 at 16:52
  • Also it looks as though the contextual information is missing; what kind of seating arrangement and how this is supposed to relate to the use of modal logic etc.
    – Mozibur Ullah
    Commented May 20, 2015 at 16:54
  • L is "necessary" operator and M is the "possible" operator.
    – MathematicalPhysicist
    Commented May 21, 2015 at 8:48

3 Answers 3

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Player-A says p is possible, exactly when player A can see someone who raises their hand for p. That means, when player-A says p is possible, then player-A can see someone.

Player A says q is necessary, exactly when everyone player-A can see raises their hand for q.

Thus if player-A can see someone, then if everyone player-A can see raises their hand for q, then player-A can see someone who raises their hand for q.

Mp -> (Lq -> Mq)

exercise 1.2: show that in any seating arrangement in which there's a player who cannot see himself Lp->p isn't valid.

My answer: Suppose A cannot see himself. Let p be on every player's sheet except A's. Suppose that A raises his hand for Lp->p then he must raise his hand for p, since he raised his hand for Lp, but he cannot see everyone's sheet (only his own), so he should keep his hand down for p, and we get that Lp->p isn't valid.

Is my reasoning ok or not?

It is not. The criteria is that the player cannot see his self.

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  • What I meant to say is that because A cannot see himself he argues that Lp is true since p is on every sheet except his own, but because p isn't on A's sheet p isn't true, thus it makes Lp->p false and thus invalid.
    – MathematicalPhysicist
    Commented May 17, 2019 at 4:47
  • Exactly, @MathematicalPhysicist . Necessary p does not entail p, since it is plausible that everyone a player sees can say p, while the player says not p (because they cannot see themselves).
    – Graham Kemp
    Commented May 17, 2019 at 5:15
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Before

and thus A raises his hand for Mq, contradiction

you could add:

and thus B raises his hand for q,

Then you have shown your formula in all seating arrangements, which is Hughes and Cresswell's way of saying "in all Kripke frames".

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From the OP:

Assume player A doesn't raise his hand for Lq ->Mq, but he raises his hand for Lq but not for Mq. (emphases added).


How can you raise your hand for Lq but not for Mq? By the rules, to raise your hand for Lq, every player you can see raised their hand for q. But the only scenario where you keep your hand down for Mq is if no one you can see raised their hand for q.

The 2nd assumption implies that both everyone (that Player A can see) raised their hand for q and yet no-one (again that Player A can see) raised their hand for q.

From the book (p. 18):

  1. If Lα is called (where α is a wff of modal logic), raise your hand if every player you can see raised his or her hand when α was called; otherwise keep your hand down.
  2. If Mα is called raise your hand if at least one of the players you can see raised his or her hand when α was called; otherwise keep your hand down.


Unfortunately the second assumption of the argument contains a contradiction.

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  • 1
    At this stage of their book (chapter 1), Hughes and Cresswell have only introduced axiom K. Without axiom D, Lq -> Mq is not a theorem, so it is not a contradiction for the player to raise their hand for La and not for Ma. It might be that the player can see nobody at all, in which case La is true and Ma is false. On the other hand, Mp -> (Lq -> Mq) is a theorem of K, because Mp requires some other player to be visible.
    – Bumble
    Commented May 16, 2019 at 23:47
  • The context of the question was the version of the game they propose, modified to account for modal operators. See p. 18. Your comment is a non-sequitur. You have NOT explained how based on the rules of the game (see my quote) how you could raise your hand for Lq but not Mq.
    – Rob
    Commented May 17, 2019 at 7:40
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    My comment is based exactly on the rules of the game. If the player can see nobody at all then they will raise their hand for every call of La, because trivially everybody they can see has raised their hand. This is why Lq -> Mq is not a theorem of K, and requires axiom D.
    – Bumble
    Commented May 17, 2019 at 12:56
  • @Bumble, I believe you are fundamentally mistaken. "If Mα is called raise your hand IF AT LEAST ONE OF THE PLAYERS you can see raised his or her hand" (p. 18). The language here seems to suggest that you must see at least one player. If you can see no one, you do not raise your hand. None does not equal some (sorry for the caps, it won't let me bold in a comment).
    – Rob
    Commented May 19, 2019 at 15:50
  • @Bumble. Also by your interpretation of the rules Player A would still have raised their hand for Mq. So you haven't explained how they could raise their hand for Lq but not Mq, which was the second assumption listed in the OP.
    – Rob
    Commented May 19, 2019 at 15:59

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