popper-logic-scientific-discovery

Now we can define,
p(a, b) = lim q(a n, b n).
This limit exists for all elements a and b of S, and it can be easily shown
to satisfy all our axioms. (See also appendix *vi, points 8 to 14.)
So much about the consistency of our axiom systems.
In order to show the independence of A1 we may take p(a, b) = 0 for
every a and b in S. Then all the axioms except A1 are satisfied.
In order to show the independence of A2 we 10 take S to consist of
three elements, S = {0, 1, 2}. We can easily show that the product ab
must be non-commutative; it may be defined as follows: 1.2 = 2; and
in all other cases, including 2.1, ab is equal to min(a, b), i.e. to the
smallest of its two components a and b. We also define: ā = 1 if and only
if a = 0; otherwise ā = 0; and we define p(0, 2) = 0; in all other cases,
p(a, b) = 1. It is now easy to show that for every b, p(1, b) = p(2, b) while
p(0, 1) = 1 and p(0, 2) = 0. Thus A2 is not satisfied. But all the other
axioms are.
We can illustrate this interpretation by writing the non-commutative
matrix as follows:
ab 0 1 2 ā
0 0 0 0 1 p(0, 2) = 0;
in all other cases
1 0 1 2 0 p(a, b) = 1
2 0 1 2 0
appendix *iv 345
In order to show that A3 is independent, we take, as in our trivial
10 In view of what has been said above about A2 it is clear that the problem of proving its
independence amounts to that of constructing an example (a matrix) which is noncommutative,
combined with a numerical rule about the p-values which ensures that the
law of commutation is violated only for the second argument. The independence proof
for A2 here described, designed to satisfy these conditions, was found at the same time
by Dr. J. Agassi and by myself. (The example satisfies Postulate AP only if in AP a bar is
placed over the letters ‘b’; but it satisfies (.) on p. 342). *Cf. Addendum on pp. 367 f.