popper-logic-scientific-discovery

The redundancy of the usual systems is due to the fact that they all
postulate, implicitly or explicitly, the validity of some or all of the rules
of Boolean algebra for the elements of S; but as we shall prove at the
end of appendix *v, these rules are all derivable from our system if we
define Boolean equivalence, ‘a = b’, by the formula
(*) a = b if, and only if, p(a, c) = p(b, c) for every c in S.
appendix *iv 349
The question may be asked whether any axioms of our system
become redundant if we postulate that ab is a Boolean product and ā a
Boolean complement; that they both obey all the laws of Boolean algebra;
and that (*) is valid. The answer is: none of the axioms (except
B1′) becomes redundant. (Only if we were to postulate, in addition, that
any two elements for which Boolean equivalence can be proved may be
substituted for each other in the second argument of the p-function, then one
of our axioms would become redundant, i.e. our axiom of substitutivity,
A2, which serves precisely the same purpose as this additional
postulate.) That our axioms remain non-redundant can be seen from
the fact that their independence (except that of A2, of course, and B1′)
can be proved with the help of examples that satisfy Boolean algebra. I
have given such examples for all except B1 and C1, for which simpler
examples have been given. An example of a Boolean algebra that shows
the independence of B1 (and A4′) and of C is this. (0 and 1 are the
Boolean zero and universal elements, and ā = 1-a; the example is,
essentially, the same as the last one, but with the probabilities −1 and 2
attached to the elements other than 0 or 1.)
ab −1 0 1 2 ā
−1 −1 0 −1 0 2
B1 (and A4′):
p(a) = a; p(a, 0) = 1;
in all other cases,
0 0 0 0 0 1 p(a, b) = p(ab)/p(b) = ab/b
C: p(a, b) = 0 if ab = 0 ≠ b;
1 −1 0 1 2 0
in all other cases,
2 0 0 2 2 −1 p(a, b) = 1.