popper-logic-scientific-discovery

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new appendices
able to show that, in the elementary theory, S is a (denumerable) Boolean
algebra. (For another example see appendix *vi, point 15.)
In Postulate 2, A1 is needed to establish that not all probabilities are equal
(say, equal to 0 or equal to 1). The function of A2 is to allow us to
prove ‘p(x, a) = p(x, b)’ for all elements a and b whose probabilities,
given any condition c, are equal. This can be done without A2, but only
under the assumption p(a) ≠ 0 ≠ p(b). Thus A2 is to enable us to extend
the probabilistic equivalence of a and b to the second argument even in
those cases in which a and b have zero absolute probability.
A2 may be replaced by the following stronger formula:
A2 +
If p(a, a) = p(b, c) = p(c, b), then p(a, b) = p(a, c);
or by either of the formulae (the weaker of which is B3 − ):
B3
B3 −
If p(ab, c) = p(ba, c), then p(c, ab) = p(c, ba).
If p(ab, ac) = p(ab, c), then p(ba, ca) = p(ba, c).
Obviously, it can therefore also be replaced by the formula (which is
simpler but much stronger):
B3 +
p(a, bc) = p(a, cb).
But since B3 + is stronger than necessary—in fact, p(a,(bc)(cb)) =
p(a, (cb)(bc)), though weaker, would suffice—it is a little misleading: its
adoption would veil the fact that with the help of the other axioms
alone, the law of commutation can be proved for the first argument.
A2 + is preferable to the other formulae here mentioned in so far as it
avoids (like the much weaker A2) using the product of a and b.
However, we can make use of the facts here stated in order to reduce
the number of our axioms to three, viz. A1, C + , and the following
axiom B which combines B3 + with B + :
B
If p(ab, c) ≠ p(a, d)p(b, c) provided that p(a, c) � p(a, d)p(b, c) and
p(a, d) = p(a, bc), then p(a, cb) ≠ p(a, d).
Apart from being stronger than one might wish it to be, this system
of only three axioms has all the advantages of the system of four
axioms A1, A2, B + , and C + .