Erratum to "Confirmation as partial entailment" [Journal of Applied Logic 11 (2013) 364-372]

Article history: Received 18 February 2014 Accepted 18 February 2014 Available online 17 March 2014 We provide a correction to the proof of the main result in Crupi and Tentori (2013). © 2014 Elsevier B.V. All rights reserved. Michael Schippers (University of Oldenburg) pointed out to us in personal correspondence an error in the proof of the main result in Crupi and Tentori [1]. The flaw spotted by Schippers is that Lemma 2 (p. 369) does not hold in its original formulation: the scheme of assignment there defined does not guarantee that one ends up with a probabilistically coherent set of values. In order to amend and validate the proof, it is sufficient to replace Lemma 2 and the subsequent lines (up to Lemma 3) by the following. Lemma 2 (Corrected). For any x, y1, y2 such that x ∈ [0, 1], y1, y2 ∈ (0, 1), there exist e, h1, h2 ∈ Lc and P ′′ ∈ P such that P ′′(h1|e)/P ′′(h1) = P ′′(h2|e)/P ′′(h2) = x, P ′′(h1) = y1, and P ′′(h2) = y2. Proof [Corrected]. Let w ∈ (0,1) be given so that w < (1− y1)/(1− xy1), (1− y2)/(1− xy2) (as the latter quantities must all be positive, w exists). The equalities in Lemma 2 arise from the following scheme of probability assignments P (h1 ∧ h2 ∧ e) = xy1y2w; P (¬h1 ∧ h2 ∧ e) = (1− xy1)xy2w; P (h1 ∧ h2 ∧ ¬e) = (1− xw)2y1y2 1− w ; P (¬h1 ∧ h2 ∧ ¬e) = [ 1− (1− xw)y1 1− w ] (1− xw)y2; P (h1 ∧ ¬h2 ∧ e) = xy1(1− xy2)w; P (¬h1 ∧ ¬h2 ∧ e) = (1− xy1)(1− xy2)w; P (h1 ∧ ¬h2 ∧ ¬e) = (1− xw)y1 [ 1− (1− xw)y2 1− w ] ; P (¬h1 ∧ ¬h2 ∧ ¬e) = [ 1− (1− xw)y1 1− w ][ 1− (1− xw)y2 1− w ] (1− w). DOI of original article: http://dx.doi.org/10.1016/j.jal.2013.03.002. * Corresponding author. http://dx.doi.org/10.1016/j.jal.2014.02.001 1570-8683/© 2014 Elsevier B.V. All rights reserved. V. Crupi, K. Tentori / Journal of Applied Logic 12 (2014) 230–231231 Suppose there exist (x, y1), (x, y2) ∈ Dk such that k(x, y1) = k(x, y2). Then, by Lemma 2 [Corrected]and the definition of Dk (see Crupi and Tentori [1, p. 369]), there exist e, h1, h2 ∈ Lc and P ′′ ∈ P such thatP(h1|e)/P (h1) = P(h2|e)/P (h2) = x, P(h1) = y1, P(h2) = y2, and P ′′(e) = w. By the probabilitycalculus, if the latter equalities hold, then P(h1 ∧ e) P(h1)P ′′(e), P(h2 ∧ e) P(h2)P ′′(e), andmoreover P(e|h1)/P ′′(e) = P(e|h2)/P ′′(e) = x. Thus, there exist e, h1, h2 ∈ Lc and P ′′ ∈ P such thateither CP ′′(h1, e) = k(x, y1) = k(x,w) = CP ′′(e, h1) even if P′′(h1 ∧ e) P′′(h1)P ′′(e), or CP ′′(h2, e) =k(x,y2) = k(x,w) = CP ′′(e, h2) even if P′′(h2 ∧ e) P′′(h2)P ′′(e), contradicting axiom A2 (see Crupi andTentori [1, p. 365]). Conversely, A2 implies that, for any (x, y1), (x, y2) ∈ Dk, k(x, y1) = k(x, y2). So, for A2to hold, there must exist a function m such that, for any e, h ∈ Lc and P ∈ P, if P (h∧ e) P (h)P (e), thenCP (h, e) = m[P (h|e)/P (h)] and m(x) = k(x, y). We then posit m : [0, 1] → and denote the domain of masDm.Up to Lemma 2 [Corrected] and then again from Lemma 3 on, the proof proceeds unchanged.1References[1] V. Crupi, K. Tentori, Confirmation as partial entailment: A representation theorem in inductive logic, J. Appl. Log. 11(2013) 364–372. 1 Meanwhile, a self-contained and corrected version of the proof is available here: http://www.vincenzocrupi.com/website/wp-content/uploads/2014/02/proof_corrected.pdf.