popper-logic-scientific-discovery

assume that (2) holds for all n, m and σ (i.e. the various arrangements
of the ones). Then there will be, according to a well-known combinatorial
theorem, n C m distinct ways of distributing m ones in n places; and
in view of the special addition theorem, we could then assert (1).
Now suppose (2) to be proved for any one n, i.e. for one particular n
and for every m and every σ which are compatible with this n. We now
show that given this assumption it must also hold for n + 1, i.e. we shall
prove
(3,0)
and
(3,1)
α F″(σ (n + 1) m + 0) = pm n + 1 − m q
α F″(σ (n + 1) m + 1) = pm + 1 (n + 1) − (m + 1) q
where ‘σ m + 0’ or ‘σ m + 1’ respectively signify those sequences of the n + 1
length which result from σ m by adding to its end a zero or a one.
Let it be assumed, for every length n of the n-tuples (or segments)
considered, that α is (at least) n − 1-free (from after-effect); thus for a
segment of the length n + 1, α has to be regarded as being at least nfree.
Let ‘σ´ m’ denote the property of being a successor of an n-tuple σ m.
Then we can assert
(4,0)
(4,1)
αF″(σ´ m.0) = αF″(σ´ m). αF″(0) = αF″(σ´ m)·q
αF″(σ´ m.1) = αF″(σ´ m). αF″(1) = αF″ (σ´ m)·p
We now consider that there must obviously be just as many σ m, i.e.
successors of the sequence ‘σ m’ in α, as there are sequences σ m in α (n),
and hence that
(5)
αF″(σ´ m) = α(n) F″(σ m)
With this we can transform the right hand side of (4). For the same
reason we have
(6,0)
αF″(σ´ m.0) = α(n + 1) F″(σ m + 0)
appendix iii 291