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Probability of being a Superset

  1. Aug 16, 2011 #1
    Let me start by saying I'm an amateur insofar as I took one propositional logic course in college and have self-taught everything else so I may be missing something obvious. However, I was reading an article by Niiniluoto (1972) and (with respect to induction) he makes the following statement:

    (2) P(h[itex]\supset[/itex]b|e) = P([itex]\neg[/itex]h[itex]\vee[/itex]b|e)

    Can someone confirm the rationale for this? I assume this is the stastical calculus version of Hempel's paradox. Specifically, the deductive version of this argument (and presumably the assumption used by Nicod and Niiniluoto) arises from the idea that:

    (a[itex]\rightarrow[/itex]b) [itex]\rightarrow[/itex] ([itex]\neg[/itex]b[itex]\rightarrow[/itex][itex]\neg[/itex]a)

    The idea being... since the second is deduced from the first, evidence for the second supports the first. However, it's trivial to create an inductive state where this is not true. Assuming this is a table of observations (column labels above, row labels to the right):

    | a | [itex]\neg[/itex]a |
    | 8 | _0 | _b__
    | 2 | _1 | [itex]\neg[/itex]b__

    In this situation

    a[itex]\rightarrow[/itex]b (80%)

    is clearly probable. However,

    [itex]\neg[/itex]b[itex]\rightarrow[/itex][itex]\neg[/itex]a (33%)

    does not follow. Intuitively, it seems to me that (in an inductive context at least) the observation of a shoe, no matter its color, can never disconfirm the hypothesis. As a consequence, it ought not confirm the hypothesis. "All" should only apply to objects/observations that can disconfirm the theory.

    What got me to the above explanation was my initial intuition that we care about the proportion of b[itex]\wedge[/itex]e that falls into h[itex]\wedge[/itex]e rather than including, for some reason, all of [itex]\neg[/itex]h including the region that is both [itex]\neg[/itex]h and [itex]\neg[/itex]b. For example, I believe what I'm saying is properly stated as:

    P(h[itex]\supset[/itex]b|e) = P(h|b[itex]\wedge[/itex]e)

    I have trouble imagining that there isn't some kind of response within Nicod's paradox for what I've described... but I can't for the life of me figure it out (or even how to look for it).

    P.S. appologize in advance for any mistakes in Latex formatting (or propositional logic)

    Ilkka Niiniluoto (1972) Inductive Systematization: Definition and a Critical Survey. Synthese, Vol. 25, No. 1/2, Theoretical Concepts and Their Operationalization (Nov. -Dec., 1972), pp. 25-81
  2. jcsd
  3. Aug 17, 2011 #2
    The fallacy here is that you're assuming that A is already true in a -> b.

    A -> B is true if ~A is true or if B is true. You're looking at A ^ B, not A -> B. Contraposition does not work for the "and" connective.
  4. Aug 17, 2011 #3

    Stephen Tashi

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    Usually probability measures are applied to sets and in most mathematical discussions when we speak of the probability of a statement or proposition, we are (according to the conventional point of view) supposed to realize that we are really speaking of a probability of a set of some sort. I don't know whether the paper that you cite is attempting to define probability in a different context.

    The proposition "if A then B" is logically equivalent to the proposition " (Not A) or B". This can be verified by comparing the truth tables for the two propositions. I've seen this fact referred to as "material implication" in logic textbooks. So there is nothing suprising about eq 2. How could the probability of two identical things not be the same?
  5. Aug 17, 2011 #4
    I spent like an hour searching the internet for the probability calculus of supersets (which is where I really got stuck) before I asked the question here. Let me take a step back and checking that this is true:

    h⊃b|e = b→h|e

    (obviously taking all the probability elements out of the statement)
  6. Aug 17, 2011 #5

    Stephen Tashi

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    My instinct is to say "yes, of course", but I suppose we should go through the formality of establishing exactly what you are talking about. The main difficulty in answering your question is interpreting your notation.

    Are the variables involved in [itex] h \supset b | e [/itex] sets or are they propositions? If they are sets, how do you interpret [itex] x \supset y [/itex] as a set when [itex] x [/itex] and [itex] y [/itex] are known sets? If they are propositions, how do you interpret " | e" ?
  7. Aug 17, 2011 #6
    Sorry, maybe i copied his stuff wrong... i'm pretty sure i follow/understand the language but am not 100% confident in its conventions... for example I thought it was safe to translate ~ to ¬ (which appears to be true) but maybe not equally , to |

    his exact formulation is

    P(h ⊃ b , e) = P( ~h ∨b , e)

    Eliminating the probabilities, I interpeted this as

    (b => h) | e = (~h ∨b) | e

    If my interpretation is right, (b => h) translates to (~b ∨ h) rather than (~h ∨b). So likely my interpretation is wrong but I need to know where so I don't repeat it.
  8. Aug 17, 2011 #7

    Stephen Tashi

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    You still didn't say whether the letters represent sets or propositions.

    Taking the letters to represents propositions ( for example "H" to mean "Merle Hedley is healthy" ) then I interpret

    "[itex] H \supset B [/itex] " to mean "[itex] H [/itex] implies [itex] B [/itex]" or, equivalently, "if [itex] H [/itex] then [itex] B [/itex]".

    The only way I see to interpret " [itex]( H \supset B) | E [/itex] as a proposition is "( [itex] H [/itex] implies [itex] B [/itex] ) and [itex] E [/itex]".

    With those interpretations, [itex]( H \supset B) | E [/itex] is logically equivalent to "((not [itex] H [/itex] ) or [itex] B[/itex]) and [itex] E [/itex]", which could be abbreviated by "[itex] (\neg H \vee B) | E [/itex]".

    People use symbols like "=", "[itex] \cup [/itex] ", "[itex] \vee[/itex] ", " [itex] \supset [/itex]" and "[itex] \neg [/itex] " both for the algebra of sets and the algebra of propositions. You have to pay attention to how the material you are reading defines its notation. Such notation is not standard across all textbooks.

    If we interpret the symbols as sets, the only interpretation that makes sense for "[itex] H \supset B [/itex]" is the set "[itex] H [/itex] complement union [itex] B [/itex]". If you try to interpret "[itex] H \supset B [/itex]" as the statement "[itex] H [/itex] is a supserset of [itex] B [/itex]" then you would be trying to apply the probability operator "[itex] P [/itex] " to a proposition on the left had side of eq 2 and apply it to a set on the right hand side of eq. 2.
  9. Aug 17, 2011 #8
    Appologize for not following the question and thanks for the clarification. While the implications are a little hazy, I at least see the difference in the context. He states:

    "all statements b, e, and h"

    ... and using his formulation of the "Deductive Theorum" plus wikipedia's version, ⊃ is → as you stated.
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