I was reading E. T. Jaynes' Confidence Intervals vs Bayesian Intervals (available here), and I came across this statement regarding Boole's The Laws of Thought:

Boole's own work on probability theory... contains ludicrous errors... See his Example 6, page 286, where by a confusion of propositions [taking the probability of the proposition: 'If X is true, Y is true' as the conditional p(Y|X)] he arrives at the conclusion that two propositions with the same truth value can have different probabilities. He not only fails to see the absurdity of this...

(the relevant section can be found here on page 228).

At first glance, I didn't understand the mistake Jaynes was referring to, but now I think it has to do with the ambiguity of saying 'the probability of if A then B'. This could either be the P(B|A) or the P(A→B), the second of which is equivalent to P(¬A ∨ B). Clearly, these probabilities are not the same, and it is easy to imagine an example case for yourself.

Boole's Mistake:

  1. 'if A then B' is equivalent to '¬A ∨ B'
  2. the probability of 'if A then B' is P(B|A)

The problem is that for (1) and (2) to be individually correct, then 'if A then B' refers to different things in each. In (1), 'if A then B' refers to the material implication. In (2), 'if A then B' refers to (B) conditioned on (A) happening.

Is my interpretation correct? Does it even make sense to talk about material implication and Bayesian conditional probabilities in the same breath?

  • 1
    Your interpretation is correct, see Nguyen et al., Probability of Implication:"There exist two different approaches to defining this probability, and these approaches lead to different probabilistic inference rules: We may interpret the probability of an implication as the conditional probability P(A|B), in which case we get Bayesian inference. We may also interpret this probability as the probability of the material implication "A or not B", in which case we get different inference rules."
    – Conifold
    Commented Apr 18 at 5:04
  • 2
    I don't think Boole is making a mistake. He shows that P(¬X ∨ Y) coincides with P(Y|X) when P(Y|X) is either one or zero. I.e. the probability of the conditional proposition of "pure Logic" as he puts it, which we now call the material conditional, agrees with the conditional probability in the case when both are certainly true or certainly false. That is not a confusion; it is correct. In practice, uncertain conditionals are nearly always correctly represented by P(Y|X) not by P(¬X ∨ Y). It is one of the limitations of the material conditional that it does not handle uncertainty correctly.
    – Bumble
    Commented Apr 18 at 10:01
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    possibly of interest: en.wikipedia.org/wiki/…
    – ac15
    Commented Apr 18 at 14:46
  • @Bumble Thank you for your response. I actually just came across your reply thread here. Would you say that Boole is misguided in thinking this means two statements with the same truth value have differing probabilities? Really, there is only one statement, the probability of which either refers to the conditional or probability of material implication. Am I just getting into semantics or is there genuine substance to this clarification?
    – adoan
    Commented Apr 18 at 15:49
  • @ac15 this is definitely beyond my familiarity with logic. I graphed the equation for the implication, and I see that when y>x the implication is certain, otherwise we have some probability, 0<=p<1, for the implication that holds whenever y=x-(p-1). Intuitively, this does not make sense to me, but I see the graph of f(x,y)=min(1,1-x+y) is the Łukasiewicz t-norm. The paper posted by a commenter above addresses my exact confusion and relates it to t-norms, so hopefully I will learn more there.
    – adoan
    Commented Apr 18 at 16:13

1 Answer 1


So example 6, on page 228 of your linked text, goes like this, translated to modern notation.

Given that P(y ∨ ¬x¬y) = p, what is P(y | x) ?

To answer this Boole introduces a constant c. He says that P(y | x) = cp / (1 - p + cp). Is this correct?

Boole describes c as the probability that "if either Y is true, or X and Y false, X is true." To me this sounds like c = P((y ∨ ¬x¬y) → x). However, the math doesn't work out with that interpretation, so this couldn't have been what Boole meant. Instead we can interpret this sentence to mean c = P(x | x → y) = P(x ∧ (x → y)) / P(x → y) = P(xy)/p.

Boole says also about c that P(x) = 1 − p + cp. This would mean c = (P(x) - 1 + p)/p, which agrees with the above interpretation.

Then Boole's formula is P(y | x) = cp / (1 - p + cp) = P(xy) / (1 - p + P(xy)). Note that 1 - p is the probability of the complement of y ∨ ¬x¬y, which is P(x¬y), so (1 - p + P(xy)) = P(x¬y) + P(xy) = P(x). So Boole's formula is equivalent to P(y | x) = P(xy) / P(x).

So Boole's formula in example 6 is correct. (To be honest I do not understand how he got there; his notation is weird.)

What about Jaynes' criticism?

he arrives at the conclusion that two propositions with the same truth value can have different probabilities.

Boole doesn't say this. To the contrary, he explains that when the two propositions have a definite truth value, then the P(y | x) and P(y ∨ ¬x¬y) coincide.

Nevertheless these expressions are such, that when either of them becomes 1 or 0, the other assumes the same value.

They only have different probabilities when P(y | x) is not definitely 1 or 0. So Jayne's criticism is misguided.

  • Thank you for your explication of Boole's reasoning. It seems that all of Boole's mathematical reasoning is correct, but is he, as Jaynes supposes, still misguided (if not wrong) in drawing the conclusion that two statements with the same truth value may have different probabilities? ie there are three probabilities in the two statements: 'if A then B' and 'B or not A'. When asking about the probability of the first statement, it is ambiguous if you mean P(A→B), which is equivalent to the probability of the second statement, or P(B|A), which is not.
    – adoan
    Commented Apr 18 at 15:44

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