popper-logic-scientific-discovery

346
new appendices
first consistency proof, S = {0, 1}, with logical products and complements
equal to the arithmetical ones. We define p(1, 1) = 1, and in all
other cases p(a, b) = 0. Then p(1, 1) ≠ p(0, 0), so that A3 fails. The other
axioms are satisfied.
In order to show that B1 is independent, we take S = { − 1, 0, + 1};
we take ab to be the arithmetical product of a and b; ā =−a; and
p(a, b) = a.(1 − |b|). Then all axioms are satisfied except B1 which
fails for a =−1, b ≠+1, and c = 0. The matrices may be written:
ab −1 0 +1 ā p(a, b) −1 0 +1
−1 +1 0 −1 +1 −1 0 −1 0
0 0 0 0 0 0 0 0 0
+1 −1 0 +1 −1 +1 0 +1 0
This example also proves the independence of A4′ (cf. note 7,
above). A second example, proving the independence of B1 and also of
B1′, is based upon the following non-commutative matrix:
ab 0 1 2 ā
0 0 1 0 2 p(0, 2) = 0;
in all other cases
1 0 1 1 0 p(a, b) = 1
2 0 1 2 0
B1 fails for a = 0, b = 1, and c = 2.
In order to show that B2 is independent, we take the same S as for
A3, and define p(0, 1) = 0; in all other cases, p(a, b) = 2. B2 fails
because 2 = p(1.1, 1) ≠ p(1, 1.1)p(1, 1) = 4, but all other axioms are
satisfied.
(Another example showing the independence of B2 can be obtained