An Observation on Carnap's Continuum and Stochastic Independencies J. B. Paris School of Mathematics The University of Manchester Manchester M13 9PL jeff.paris@manchester.ac.uk Abstract We characterize those identities and independencies which hold for all probability functions on a unary language satisfying the Principle of Atom Exchangeability. We then show that if this is strengthen to the requirement that Johnson's Sufficientness Principle holds, thus giving Carnap's Continuum of inductive methods for languages with at least two predicates, then new and somewhat inexplicable identities and independencies emerge, the latter even in the case of Carnap's Continuum for the language with just a single predicate. Keywords: Carnap's Continuum, Atom Exchangeability, Stochastic Independence, Inductive Logic. Introduction To this day the Continuum of Inductive Methods described by Carnap in [1], [2], [3], [4] continues to be adapted and promoted as paradigm solutions to various problems within Inductive Logic. For example arithmetic combinations of these functions figure almost exclusively in recent attempts to provide probability functions exhibiting certain specific features of analogical influence, see [5], [11], [12], [15], [16]. There seem to be several good reasons for this focus. Firstly this Continuum has a widely acceptable justification in terms of its 'rationality': There is a putatively rational requirement, namely Johnson's Sufficientness Principle, that we can impose on an inductive method (i.e. probability function) which Preprint submitted to Applied Logic February 3, 2017 forces it to be precisely a member of Carnap's Continuum (see also Johnson's earlier derivation of this in [9]), at least when we assume that the language has more than one predicate. Secondly the Continuum has a simple form, making it easy to work with, whilst the parameter it involves has a clear interpretation which readily permits generalizations. Carnap's original goal in his Inductive Logic programme was to develop an inductive method which could be applied to real world problems of induction, or more generally the assignment of probabilities based on some finite body of evidence, and which furthermore was logical in the sense that it's conclusions followed mechanically from the evidence via certain precisely formulated rules or principles. The arrival on the scene of Goodman's Grue Paradox, [6], [7], however highlighted an evident flaw in the practicality of the approach; that in real (as opposed to toy) examples there is usually so much available evidence that even if it could be suitably formulated in the language of the problem it would be completely infeasible to take it as one's premise set. Whilst many philosophers have seen this as the end of the programme as a practical, rather than simply a theoretical, project, nevertheless apparently similar aspirations to Carnap's still seem to underlie papers such as those on analogical reasoning cited above. One explanation for this is that whilst all our available knowledge in an real world situation is just too much to handle nevertheless most of it should be redundant or irrelevant and possibly what really does matter can be simply formulated. This raises the question we shall consider in this paper, to what extent is this a reasonable assumption for the members of Carnap's Continuum, more precisely under what circumstances is a sentence θ stochastically independent of a sentence φ for all members of Carnap's Continuum? Before that however we need to spend a little time introducing some standard notation. The experienced reader might therefore be advised to skip the next section, only referring back to it as necessary. Notation Let L be a predicate language with just q (unary) predicates, P1, P2, . . . , Pq, constants a1 for i = 1, 2, 3, . . . and no other relation, constant or function symbols. As usual the intention here is that these ai exhaust the universe. Let α1(x), α2(x), . . . , α2n(x) denote the atoms of L, that is the 2 q formulae 2 of L of the form ±P1(x) ∧ ±P2(x) ∧ . . . ,∧ ± Pq(x). So for example the atoms in the case q = 2 are P1(x)∧P2(x), P1(x)∧¬P2(x), ¬P1(x) ∧ P2(x), ¬P1(x) ∧ ¬P2(x). Knowing which atom an ai satisfies tells us exactly which of the Pj(x) ai does or does not satisfy, and hence tells us everything there is to know about ai. A state description, Θ(b1, b2, . . . , bm), for distinct choices b1, b2, . . . , bm from the ai, is a sentence of the form m∧ i=1 αji(bi), (1) and similarly tells us all there is to know about b1, b2, . . . , bm. Notice that the state descriptions for b1, b2, . . . , bm are disjoint and any quantifier free sentence φ(b1, b2, . . . , bm) of L is logically equivalent to a disjunction s∨ k=1 Θk(b1, b2, . . . , bm) of distinct state descriptions Θk(~b) for b1, b2, . . . , bm. Hence if w is a probability function on L (for a definition see for example [8] or [14]) then w(φ(b1, b2, . . . , bm)) = s∑ k=1 w(Θk(b1, b2, . . . , bm)). (2) We say that w satisfies Constant Exchangeability, Ex, if w(φ(b1, b2, . . . , bm)) depends only on φ(x1, x2, . . . , xm) and not on the (distinct) instantiating constants b1, b2, . . . , bm. By (2) it is already enough that this holds for state descriptions. Since all the probability functions we shall consider will satisfy Ex our results will apply for general b1, b2, . . . , bm once proven for a1, a2, . . . , am. The spectrum of a state description Θ(b1, . . . , bm) as in (1) is the multiset 1 n = {n1, n2, . . . , n2q }, where ni is the number of times that the atom αi(x) appears amongst the αj1(x), αj2(x), . . . , αjm(x). 1Multisets are just like sets except that elements may be repeated. 3 We say that w satisfies Atom Exchangeability, Ax, if w(Θ(b1, b2, . . . , bm) depends only on the spectrum n of the state description Θ(b1, b2, . . . , bm). In this case we shall write w(n) for w(Θ(~b)). Finally we say that w satisfies Johnson's Sufficientness Principle, JSP, if for a state description Θ(b1, b2, . . . , bm) as in (1), w(αi(bm+1) |Θ(b1, b2, . . . , bm)) depends only on m and ni. It is well known that JSP implies Ax which in turn implies Ex. As shown originally by Johnson, [9] (and independently later by Kemeny, see [4, section 19] and [10]) if the number of predicates, q, is at least 2 and the probability function w satisfies JSP then w is a member of Carnap's Continuum of Inductive Methods. That is, w = cλ for some 0 ≤ λ ≤ ∞ where, with the above notation, cλ is the probability function satisfying Ax such that cλ((αi(bm+1) ∧Θ(b1, b2, . . . , bm)) = (ni + λ/2 q) (m+ λ) * cλ(Θ(b1, b2, . . . , bm)). (3) The cases λ = 0,∞ here are rather exceptional and until further notice we shall restrict ourselves to 0 < λ < ∞ when discussing the cλ (though still referring to these as Carnap's Continuum.) Stochastic Independence and Ax Let w be a probability function on L satisfying Ax. Then from (2) for φ(a1, . . . , am) a sentence of L, w(φ(a1, . . . , am)) = ∑ n fφ(n)w(n), where the n range over the possible spectra {n1, n2, . . . , n2q } with ∑2q i=1 ni = m and fφ(n) is the number of state descriptions in (2) with spectrum n. Hence the stochastic independence2 of θ(a1, . . . , am) and φ(a1, . . . , am) with respect to w, i.e. w(θ(a1, . . . , am) ∧ φ(a1, . . . , am)) = w(θ(a1, . . . , am)) * w(φ(a1, . . . , am)) 2If w(φ) 6= 0 we may replace this condition by the equivalent w(θ |φ) = w(θ) if in the context that seems more appropriate. 4 amounts to the identity,(∑ n fθ∧φ(n)w(n) ) = (∑ n fθ(n)w(n) ) * (∑ n fφ(n)w(n) ) , equivalently,(∑ n fθ∧φ(n)w(n) ) * (∑ n f>(n)w(n) ) = (∑ n fθ(n)w(n) ) * (∑ n fφ(n)w(n) ) (4) since ∑ n f>(n)w(n) = 1. We can now turn (briefly as it happens) to the main question we are interested in: When does (4) hold for cλ with 0 < λ <∞ ? Given the aspirations outlined in the first section one may hope that we should certainly have independence when θ(~a) and φ(~a) are respectively logically equivalent to sentences θ′(a1, a2, . . . , ar) and φ ′(ar+1, ar+2, . . . , am) which have no predicates nor constants in common. However, as already pointed out in [8]3, this can fail, for example for 0 < λ <∞, cλ(P2(a3) ∧ P2(a4) |P1(a1) ∧ P1(a2)) > cλ(P2(a3) ∧ P2(a4)). Given this observation one might rashly be inclined to quite the opposite view, that the cλ do not satisfy any independencies such as (4) except in the trivial cases when ±θ ≡ > or ±φ ≡ >. This is not true however in at least two ways. Firstly the identity (4) with w = cλ is equivalent to a polynomial identity in the variable λ which will sometimes hold for a finite set of roots λ. We will dispense with such chance independencies by considering which independencies of the form (4) hold for all the cλ (with 0 < λ <∞). There is however a second way in which (4) can hold for all the cλ. For if q ≥ 2 then the f(n) will have common divisors and simply by taking disjunctions of 3On the other hand there is an argument why this inequality is desirable: Namely the P1(a1)∧P1(a2) provides evidence that the individuals ai are similar and hence should support the view that a3, a4 are similar, in particular positively supporting P2(a3)∧P2(a4). 5 state descriptions we can construct sentences θ(a1, a2, . . . , am), φ(a1, a2, . . . , am) such that for some constant k, with 0 < k < 1, and each n fθ∧φ(n) = kfφ(n), fθ(n) = kf>(n). In this case the equality (4) will of course hold for all probability functions w satisfying Ax. For that reason we shall now temporarily break off from the main interest of this paper and consider the question of what identities and independencies must hold for all probability functions satisfying Ax.4 In fact Theorem 3 below will show that the independencies described above are exactly those which hold for all probability functions w satisfying Ax. First though we will derive some special cases which are perhaps of independent interest. Proposition 1. If equation (4) holds for all probability functions w on L satisfying Ax then either ∑ n fθ(n)w(n) is constant for all probability functions w satisfying Ax or∑ n fφ(n)w(n) is constant for all probability functions w satisfying Ax. Proof Suppose that (4) held for every probability function w satisfying Ax but that for two such functions w1, w2,∑ n fθ(n)w1(n) 6= ∑ n fθ(n)w2(n). (5) In this case θ cannot be logically equivalent to ⊥ and by small perturbations of w1, w2 if necessary we may assume that neither side of (5) is zero. 4It is easy to see that if we weaken Ax to Ex then the only independencies satisfied by all probability functions satisfying Ex are those of the form w(θ ∧ φ) = w(θ) * w(φ) when ±θ ≡ > or ±φ ≡ >. 6 The probability function (w1 +w2)/2 will also satisfy Ax and if according to wi (i = 1, 2) (4) gives Ai = BiCi where Ai = ∑ n fθ∧φ(n)wi(n), Bi = ∑ n fφ(n)wi(n), Ci = ∑ n fθ(n)wi(n) then according to (w1 + w2)/2 (4) gives 2(A1 + A2) = (B1 +B2)(C1 + C2). If B1B2 = 0 this together with (5) gives B1 = B2 (= 0). Otherwise multiplying by B1B2 and eliminating the C1, C2 gives (A1B2 − A2B1)(B1 −B2) = 0 and again with (5) it follows that B1 = B2. In other words the second possibility in the proposition pertains for w1 and w2. Now if the proposition failed there would be w1, w2, w3, w4 such that∑ n fθ(n)w1(n) 6= ∑ n fθ(n)w2(n), ∑ n fφ(n)w1(n) = ∑ n fφ(n)w2(n),∑ n fθ(n)w3(n) = ∑ n fθ(n)w4(n),∑ n fφ(n)w3(n) 6= ∑ n fφ(n)w4(n), with all such quantities non-zero. But by applying the same argument as above for (w1 +w3)/2 and (w2 +w4)/2 we see that this is not possible. The result follows.  Proposition 2. With the notation of the previous proposition suppose that for all probability functions w satisfying Ax that∑ n fθ(n)w(n) = k (6) for some constant k. Then for each n, fθ(n) = kf>(n). 7 Proof Given reals s1, s2, . . . , s2q ≥ 0 and not all zero let w~s be the probability function on L such that w~s(n) = (2 q!)−1 ∑ σ sn1σ(1)s n2 σ(2) . . . s n2q σ(2q)(s1 + s2 + . . .+ s2q) −m where σ ranges over the permutations of 1, 2, . . . , 2q. Then w~s satisfies Ax and (6) gives that∑ n fθ(n)(2 q!)−1 ∑ σ sn1σ(1)s n2 σ(2) . . . s n2q σ(2q) = k(s1 + s2 + . . .+ s2q) m. Since we can take each si to be algebraically independent this is only possible if the coefficients of sn11 s n2 2 . . . s n2q 2q on both sides agree, from which the result follows.  Theorem 3. Equation (4) holds for all probability functions w on L satisfying Ax if and only if ±θ ≡ > or ±φ ≡ > or for some constant k, 0 < k < 1, fθ∧φ(n) = kfφ(n), fθ(n) = kf>(n) for all n (7) or for some constant k, 0 < k < 1, fθ∧φ(n) = kfθ(n), fφ(n) = kf>(n) for all n. (8) Proof Assume that θ, φ are neither tautologies nor contradictions. In the first case of Proposition 1 we may assume that∑ n fθ(n)w(n) = k (9) for some constant k, with 0 < k < 1, and for all probability functions w satisfying Ax. Hence from (4),∑ n fθ∧φ(n)w(n) = k ∑ n fφ(n)w(n). Now using w~s as in the proof of Proposition 2 we obtain that fθ∧φ(n) = kfφ(n) for all n and since we already have fθ(n) = kf>(n) for all n, again by Proposition 2 with (9), the result (7) follows. 8 In the second case ∑ n fφ(n)w(n) = k for some constant k, with 0 < k < 1 and (8) follows analogously. The converse is of course immediate from our earlier observations.  We remark that by utilizing w~s with each of the si are algebraically independent we obtain probability functions which satisfy Ax and whose only independencies are those which all probability functions satisfying Ax must satisfy. Stochastic Independence and JSP We now return again to considering those non-trivial identities cλ(θ(~a) ∧ φ(~a)) = cλ(θ(~a)) * cλ(φ(~a)) which hold for all 0 < λ < ∞, where now 'non-trivial' means not holding for all probability functions satisfying Ax. It turns out that there are many such independencies but first we prove a negative result, recalling that m is the number of constants mentioned in θ(~a) and φ(~a): Theorem 4. For m ≤ 3, equation (4) holds for all probability functions cλ in Carnap's Continuum if and only if it holds for all probability functions w satisfying Ax. Proof It is enough to prove the result for m = 3. For (4) to hold for cλ with θ(~a), φ(~a) non-contradictory requires that for f1 = fθ∧φ({1, 1, 1}), f2 = fθ∧φ({2, 1}), f3 = fθ∧φ({3}), h1 = fθ({1, 1, 1}), g1 = fφ({1, 1, 1}), t1 = f>({1, 1, 1}) etc., (f1μ 2 + f2μ(μ+ 1) + f3(μ+ 1)(μ+ 2))(t1μ 2 + t2μ(μ+ 1) + t3(μ+ 1)(μ+ 2)) = (h1μ 2+h2μ(μ+1)+h3(μ+1)(μ+2))(g1μ 2+g2μ(μ+1)+g3(μ+1)(μ+2)) (10) where μ = λ/2q and none of these polynomials is identically zero. Let f = f1 + f2 + f3 etc.. Clearly if (10) is to hold for all μ then (f/h) = (g/t). Factorizing the polynomial factors in (10) gives, say, (f(μ+ ζ1)(μ+ ζ2))(t(μ+ δ1)(μ+ δ2)) = 9 (h(μ+ γ1)(μ+ γ2))(g(μ+ β1)(μ+ β2)). (11) There are now various possibilities: (a) {ζ1, ζ2} = {γ1, γ2}. In this case a similar phenomenon must hold for the β, δ and, since the polynomials μ2, μ(μ + 1), (μ + 1)(μ + 2) are linearly independent, this gives fθ∧φ(n) = (f/h)fθ(n), fφ(n) = (f/h)f>(n), and so this independency holds for all probability functions satisfying Ax. (b) {ζ1, ζ2} = {β1, β2}. This case follows as in the previous case. (c) Not cases (a) or (b). Notice that in this case the ζ1, ζ2, β1 etc. must all be real since otherwise ζ1, ζ2 must be conjugates, etc. and one of the previous cases must have held. But since t1 = 2 q(2q − 1)(2q − 2), t2 = 3 * 2q(2q − 1), t3 = 2q, t(μ+ δ1)(μ+ δ2) = tμ 2 + (t2 + 3t3)μ+ 2t3 = 2 q((2qμ)2 + (2qμ) + 2) which has complex roots, so this case cannot occur and the required result follows.  However for q ≥ 2 the situation changes once m > 3. Proposition 5. For m > 3 and q ≥ 2 there are identities of the form (4) which hold for all probability functions cλ in Carnap's Continuum but fail for some probability function w satisfying Ax. Proof Take q = 2 and the usual atoms α1(x) = P1(x) ∧ P2(x), α2(x) = P1(x) ∧ ¬P2(x), α3(x) = ¬P1(x) ∧ P2(x), α4(x) = ¬P1(x) ∧ ¬P2(x). In this case one can check that 2cλ(α 2 1α 2 2) = cλ(α 2 1α2α3) + cλ(α 3 1α2), (12) where α21α 2 2 is short for α1(a1) ∧ α1(a2) ∧ α2(a3) ∧ α2(a4) etc.. Writing ~a for a1, a2, a3, a4 let φ(~a) = (P2(a2) ∧ ¬(Θ1(~a) ∨Θ2(~a))) ∨Θ3(~a) ∨Θ4(~a) 10 where Θ1(~a),Θ2(~a),Θ3(~a),Θ4(~a) are respectively the state descriptions α1(a1) ∧ α1(a2) ∧ α1(a3) ∧ α4(a4), α1(a1) ∧ α1(a2) ∧ α2(a3) ∧ α4(a4), α1(a1) ∧ α2(a2) ∧ α2(a3) ∧ α1(a4), α1(a1) ∧ α2(a2) ∧ α1(a3) ∧ α2(a4). In this case cλ(P1(a1)) = 1/2 = cλ(P1(a2)) and by counting contributing state descriptions for ~a of a particular spectrum n we see that when n = { 2, 2 }, { 2, 1, 1 }, { 3, 1 }, { 4 }, { 1, 1, 1, 1 }, P1(a1) has 18, 72, 24, 2, 12 such respectively, as does P2(a2). If we were to take φ(~a) = P2(a2) we would obtain corresponding figures of 9, 36, 12, 1, 6 for P1(a1) ∧ φ(~a). However if we just remove two state descriptions from P1(a2) and add two extra ones as in the φ(~a) defined above the corresponding figures for φ(~a) and P1(a1)∧φ(~a) come out to be 20, 71, 23, 12, 2 and 11, 35, 11, 6, 1. Using (12) it now follows that cλ(P1(a1) ∧ φ(~a)) = cλ(P1(a1)) * cλ(φ(~a)). However we can certainly find probability functions w satisfying Ax for which w(P1(a1) ∧ φ(~a)) 6= w(P1(a1)) * w(φ(~a)). For example let wδ be as in the Nix-Paris Continuum, see [13], and 0 < δ < 1. In this case for ν = (1 + 3δ)/(1− δ) and C = 4−5(1− δ)4, wδ({ 2, 2 }) = 2C(ν2+1), wδ({ 2, 1, 1 }) = C(ν+1)2, wδ({ 3, 1 }) = C(ν3+ν+2) wδ({ 4 }) = C(ν3 + 3), wδ({ 1, 1, 1, 1 }) = 4Cν, and wδ(φ(~a)∧P1(a1)) = C(11(2ν2+2)+35(ν+1)2+11(ν3+ν+2)+6(4ν)+(ν3+3)) which is not in general the same as wδ(φ(~a)) * wδ(P1(a1)) = 2−1C(18(2ν2 + 2) + 72(ν+ 1)2 + 24(ν3 + ν+ 2) + 12(4ν) + 2(ν3 + 3)).  11 We shall delay further discussion of this example until the next section. Right now we will consider the case when q = 1, that is when our language only has a single predicate, P say. In this case we can show Theorem 5 also for m = 4. When m = 5 we again return to the situation of Proposition 5, though the same method will not adapt. That is, when we only have a single predicate we cannot utilize a non-trivial identity of the form cλ(ψ(~a)) = kcλ(η(~a)) for some constant k to construct θ(~a), φ(~a) such that for all 0 < λ <∞ cλ(θ(~a) ∧ φ(~a)) = cλ(θ(~a)) * cλ(φ(~a)), whilst this identity fails for some probability function w satisfying Ax. In more detail: Theorem 6. Suppose that L has only a single predicate (i.e. q = 1) and cλ(ψ(~a)) = kcλ(η(~a)) (13) for all 0 < λ < ∞. Then fψ(n) = kfη(n) for all n and hence the identity (13) holds for all probability functions satisfying Ax. Proof Let x = λ/2 and set gn(x) = n−1∏ j=0 (x+ j), so g0(x) = 1. Notice that cλ(Θ(a1, . . . , am)) = gn(x)gm−n(x)∏m−1 j=0 (j + 2x) = gn(x)gm−n(x) gm(2x) for Θ(~a) a state description having spectrum {n,m− n}. For 0 ≤ n ≤ m, g[m/2](x) | gn(x)gm−n(x) where as usual [m/2] is the integer part of m/2. Let qn(x) = gn(x)gm−n(x) g[m/2](x) . 12 for n ≤ m/2. By considering the values of q0(x), q1(x) . . . , q[m/2] at 0,−1,−2, . . . , −[m/2] + 1 it follows that these qn(x) are linearly independent. Since for q = 1 the maximum length of a spectrum for this language is 2, cλ(ψ(~a)) = g[m/2](x) ∑ n≤[m/2] fψ({n,m− n})qn(x), cλ(η(~a)) = g[m/2](x) ∑ n≤[m/2] fη({n,m− n})qn(x). From (13) we must have g[m/2](x) ∑ n≤[m/2] fψ({n,m− n})qn(x) = kg[m/2](x) ∑ n≤[m/2] fη({n,m− n})qn(x) so using the above linear independencies we must have fψ({n,m− n}) = kfη({n,m− n}) for n ≤ [m/2] and the result follows.  An alternative approach to that given in the proof of Proposition 5 does however give a non-trivial independency for m = 5 and q = 1 as we now show. Proposition 7. Suppose that L has only a single predicate (i.e. q = 1). Then there are θ(a1, . . . , a5), φ(a1, . . . , a5) such that for all 0 < λ <∞ cλ(θ(~a) ∧ φ(~a)) = cλ(θ(~a)) * cλ(φ(~a)) but this fails for some probability function w satisfying Ax. Proof Using the notation of the proof of the previous theorem, for m = 5 g[m/2](x) = x(x+ 1)(x+ 2), g0(x)g5(x) = x(x+ 1)(x+ 2)(x+ 3)(x+ 4), q0(x) = (x+ 3)(x+ 4), g1(x)g4(x) = x 2(x+ 1)(x+ 2)(x+ 3), q1(x) = x(x+ 3), g2(x)g3(x) = x 2(x+ 1)2(x+ 2), q2(x) = x(x+ 1), 13 and 2q0(x) + 10q1(x) + 20q2(x) = 8(2x+ 1)(2x+ 3), q0(x) + 4q1(x) + 15q2(x) = 2(5x+ 6)(2x+ 1), 4q1(x) + 12q2(x) = 8x(2x+ 3), q1(x) + 9q2(x) = 2x(5x+ 6). Omitting mention of ~a = 〈a1, a2, a3, a4, a5〉 let Θ1,Θ2 be the two state descriptions with spectrum {5}, let Φ1, . . . ,Φ10 be the 10 state descriptions with spectrum {1, 4}, let Ψ1, . . . ,Ψ20 be the 20 state descriptions with spectrum {2, 3} and let φ(a1, . . . , a5) = Θ1 ∨ 4∨ i=1 Φi ∨ 15∨ i=1 Ψi, θ(a1, . . . , a5) = 7∨ i=4 Φi ∨ 18∨ i=7 Ψi, ψ(a1, . . . , a5) = Φ4 ∨ 15∨ i=7 Ψi. Then φ(~a) ∧ θ(~a) ≡ ψ(~a) and cλ(ψ(~a) |φ(~a)) = q1(x) + 9q2(x) q0(x) + 4q1(x) + 15q2(x) = 2x(5x+ 6) 2(5x+ 6)(2x+ 1) , cλ(θ(~a) | >) = 4q1(x) + 12q2(x) 2q0(x) + 10q1(x) + 20q2(x) = 8x(2x+ 3) 8(2x+ 1)(2x+ 3) , so cλ(θ(~a) ∧ φ(~a)) = cλ(θ(~a)) * cλ(φ(~a)). However by Theorem 1 this identity is clearly not trivial in the sense of being satisfied by all probability functions w satisfying Ax.  Discussion The previous sections have shown that there are in fact many 'mysterious' identities and independencies which hold for all the cλ in Carnap's Continuum, since, when suitably phrased the results for 0 < λ <∞ can be extended 14 to include also λ = 0,∞ by a continuity argument. These are 'mysterious' in the sense that they do not hold for all probability functions satisfying Ax, if they did then by Theorems 2, 3, they would be easily explained and comprehended. Instead their derivation must require the stronger assumption of JSP rather than just Atom Exchangeability. However whilst the content of Johnson's Sufficientness Principle appears easy to grasp, and for the sake of argument accept, this seems not at all to be the case for these mysterious consequences. For example from the identity (12) and Atom Exchangeability we can obtain that cλ((P2(a1)↔ ¬P2(a2)) ∧ P2(a3) ∧ ¬P2(a4) ∧ 4∧ i=1 P1(ai)) = cλ(P2(a1)∧P2(a2)∧¬P1(a4)∧¬P2(a4)∧ 3∧ i=1 P1(ai)) (14) whilst from Proposition 5 we have that cλ(P1(a1) |φ(a1, a2, a3, a4)) = 1/2 = cλ(P1(a1)), where φ(~a) is (the somewhat incomprehensible) [P2(a2) ∧ ¬(P1(a3) ∧ ¬P1(a3) ∧ ¬P2(a4) ∧ 2∧ i=1 (P1(ai) ∧ P2(ai))] ∨ [P2(a1) ∧ ¬P2(a2) ∧ (P2(a3)↔ ¬P2(a4)) ∧ 4∧ i=1 (P1(ai)]. The immediate conclusion this leads us to then seems to be that there is much more that is hidden and mysterious in Johnson's Sufficientness Principle, and in turn Carnap's Continuum, than we might have expected. Whether or not one can give an enlightening explanation which will dispel the fog remains to be seen. 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