Proof of the reflexivity of the accessibility relation

If accessibility is defined as follows:

Definition 1. wRw' ↔ ∀p (w' ⊨ p → w ⊨ ◊p)

and the following axioms hold:

Axiom 1. p → ◊p,

Axiom 2.pw (w ⊨ p → w ⊨ ◊p),

are the following claims true?

Claim 1.w (w ⊨ p → w ⊨ ◊p) (by Axiom 1 and Definition 1),

Claim 2.w (wRw) (by Axiom 2 and Claim 1).

The task is to prove the correspondence between the so-called T axiom (φ → ♢φ) and the reflexivity of the accessibility relation. To do so we must be very explicit about the following important distinction:

Definition 1a. (Frame) A frame F = (W, R) is a directed graph.

Definition 1b. (Model) A model M = (F, V) is a labelled directed graph.

Directed graphs are simply a bunch of connected points (in this context, the points are called 'worlds', the connections form the 'accessibility relation'). Labelled directed graphs are directed graphs but with each point having a label that tells us which propositions are true at that point. We want to show that:

Fact 2. Formula (φ → ♢φ) is valid iff R is reflexive.

It turns out, however, that "[t]ruth in models is not appropriate for bringing out such correspondences" between formulas and properties of the accessibility relation, "as special valuations [in our terminology: labelings] may validate axioms even though the underlying frame has no nice behavior at all" (van Benthem 2010, p. 101). So what we really want to show is that:

Fact 2. Formula (φ → ♢φ) is valid on frame F = (W, R) iff R is reflexive.

The definition of frame validity is simple to formulate:

Definition 3. (Frame Validity) Formula φ is valid on a frame F = (W, R) iff for all models M = (F, V) based on that frame and worlds w ∈ W, we have (M, w) |= φ.

The "for all models" is the key. Let's now tackle the task by splitting it into its two directions:

Fact 2a. (Only if direction) Formula (φ → ♢φ) is valid on frame F = (W, R) only if R is reflexive.

Proof. We prove by contraposition. Assume that F = (W, R) is not reflexive, with the goal of showing that (φ → ♢φ) is not valid on F. To show that (φ → ♢φ) is not valid on F it will suffice to present a counter-model M = (F, V) based on F where for some w ∈ W, (M, w) |= φ ∧ ¬♢φ. Consider M = (W, R, V), where W = {w}, R is irreflexive and V(p) = W. Informally, we have a single node called 'w' with the label 'p' meaning that proposition p is true at world w; we have no other worlds. Now, since p is true at w, we have (M, w) |= p. But since there are no other worlds and (by irreflexivity of R) w cannot access itself, (M, w) |= ♢p is false. That means that we have a pointed world (M, w) s.t. (M, w) |= p ∧ ¬♢p. That means that (φ → ♢φ) cannot be valid on frame F.                                                                                 ■

Fact 2b. (If direction) Formula (φ → ♢φ) is valid on frame F = (W, R) if R is reflexive.

Proof. We assume that R is reflexive, with the goal of showing that (φ → ♢φ) is valid on reflexive frames. Let M = (F, V) be any model based on such a frame. For an arbitrary point w ∈ W, we want to show that (M, w) |= φ → ♢φ. Suppose, for contradiction, that (M, w) |= φ → ♢φ is false. That will be the case iff (M, w) |= φ and (M, w) |= ¬♢φ are both true. By the box-diamond duality, this second conjunct is true just in case (M, w) |= ▢¬φ is true. But by the first conjunct, world w satisfies φ and since R is reflexive, it means that w sees a world (namely itself) that satisfies φ, so (M, w) |= ▢¬φ cannot be true. Since the assumption that (M, w) |= φ → ♢φ is false led to a contradiction, we know that (M, w) |= φ → ♢φ is true. Since (M, w) was arbitrary, we know that (φ → ♢φ) is valid on reflexive frames.                ■

Those two facts allow us to justify the desired claim, namely that:

Fact 2. Formula (φ → ♢φ) is valid on frame F = (W, R) iff R is reflexive.

Proof. Combine proofs of Fact (2a) and (2b).                                                                                          ■

References

van Benthem, J. (2010) Modal Logic for Open Minds, Stanford, CSLI Lecture Notes #199; §9.1.
Holliday, W.H. (2012) Modal Reasoning, Lecture Course (Spring), UC Berkeley; Lecture 5.
Pacuit, E. (2009) "Notes on Modal Logic"; §3.1.

• On Fact 2a. you say "R is reflexive and V(p) = W" V(p) = W means p is true in/at W, correct? Sorry if that sound naïve, i'm still a beginner in (modal) logic. Aug 5 '14 at 20:47
• @user8083 Exactly (since W = {w}, V(p) = W = {w}, i.e. p is true at w). Aug 5 '14 at 20:52
• @user8083 I hope after looking at the answer to your other question this now makes sense. Let me know if you have any questions. Aug 5 '14 at 21:53
• What good introductory textbook(s) on modal logic would you recommend? I've tried using Modal Logic for Open Minds, but it was too technical. Aug 6 '14 at 0:03
• @user8083 Hughes and Cresswell I hear is pretty accessible. Van Benthem has an earlier book on modal logic that's a little more accessible. Take a look at Pacuit's Notes I listed in the references section. John MacFarlane has excellent introductory notes on his website. I'll think more about this and comment when I have better suggestions. Aug 6 '14 at 0:18