popper-logic-scientific-discovery

364
new appendices
In order to prove now that, in any consistent interpretation, S must
be a Boolean algebra, we note that
(98) ((x)p(a, x) = p(b, x)) → p(ay, z) = p(by, z) B2
(99) ((x)p(a, x) = p(b, x)) → p(y, az) = p(y, bz) 98, A2
It is interesting that (99) needs A2: it does not follow from 98, 40, and
B2, since it is possible that p(a, z) = p(b, z) = 0. (This will be the case, for
example, if ā = z ≠ xx¯.)
(100) ((x)(p(a, x) = p(b, x) & p(c, x) = p(d, x))) → p(ac, y) = p(bd, y)
99, B2
With the help of (90), of (100), and of A2, it can now easily be
shown at once that whenever the condition
(*)
p(a, x) = p(b, x) for every x in S
is satisfied, any name of the element a may be substituted for some or
all occurrences of names of the element b in any well-formed formula
of the calculus without changing its truth value; or in other words, the
condition (*) guarantees the substitutional equivalence of a and b.
In view of this result, we now define the Boolean equivalence of two
elements, a and b, as follows.
(D1) a = b ↔ (x)p(a, x) = p(b, x)
From this definition we obtain at once the formulae
(A) a = a
(B) a = b → b = a
(C) (a = b & b = c) → a = c
(D) a = b → a may replace b in some or all places of any formula
without affecting its truth value. A2, 90, 100
We may also introduce a second definition
(D2) a = b + c ↔ a = b c¯