In [DarwPearl:AIJ], Darwiche and Pearl propose postulates for iterated revisions, noted (R*1) to (R*6) and (C1) to (C4). In particular, the postulate (C3) reads:

(C 3) I f Ψ \circ α ⊨ μ, t h e n (Ψ \circ μ) \circ α ⊨ μ .

It will be shown that, in the presence (R*1) to (R*6), (C1) and (C3) imply:

(* *) I f Ψ \circ α ⊨ μ, t h e n (Ψ \circ μ) \circ α \equiv Ψ \circ α .

First, a lemma.

Lemma 1

Assuming (R*1) to (R*6), if $Ψ \circ μ ⊨ φ$ , then $Ψ \circ μ \equiv Ψ \circ (μ \land φ)$ .

Proof: Since $Ψ \circ μ ⊨ φ$ , $Ψ \circ μ ⊨ (Ψ \circ μ) \land φ$ . By (R*4), $(Ψ \circ μ) \land φ ⊨ Ψ \circ (μ \land φ)$ . Therefore $Ψ \circ μ ⊨ Ψ \circ (μ \land φ)$ .

If $Ψ \circ μ$ is satisfiable, then, since $Ψ \circ μ ⊨ φ$ , $(Ψ \circ μ) \land φ$ is satisfiable and, by (R*5), $Ψ \circ (μ \land φ) ⊨ (Ψ \circ μ) \land φ$ and therefore $Ψ \circ (μ \land φ) ⊨ Ψ \circ μ$ .

If $Ψ \circ μ$ is not satisfiable, then, by (R*3), $μ$ is not satisfiable, and $μ \land φ$ is not satisfiable. By (R*1), then, $Ψ \circ (μ \land φ) ⊨ Ψ \circ μ$ .

Lemma 2

Assuming (R*1) to (R*6), (C1) and (C3), if $Ψ \circ α ⊨ μ$ , then $(Ψ \circ μ) \circ α \equiv Ψ \circ α$ .

Proof: Suppose $Ψ \circ α ⊨ μ$ . By Lemma 1, $Ψ \circ α \equiv Ψ \circ (α \land μ)$ . By (C1), $Ψ \circ (α \land μ) \equiv (Ψ \circ μ) \circ (α \land μ)$ . But, by (C3), $Ψ \circ μ \circ α ⊨ μ$ and, by Lemma 1, $Ψ \circ μ \circ α \equiv Ψ \circ μ \circ (α \land μ)$ .

We conclude that $Ψ \circ α \equiv Ψ \circ μ \circ α$ .

A note on Darwiche and Pearl

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1 A short remark

Lemma 1

Lemma 2

References