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Rh with the cardinal number $$\nu+1$$. If the theorem is true for this aggregate, its truth for any other aggregate with the same cardinal number $$\nu+1$$ follows at once by § 1. Let $$E'$$ be any part of $$E_\nu$$; we distinguish the following cases:

(a) $$E'$$ does not contain $$e_\nu$$ as element, then $$E$$ is either $$E_{\nu-1}$$ [491] or a part of $$E_{\nu-1}$$, and so has as cardinal number either $$\nu$$ or one of the numbers $$1, 2, 3, ..., \nu-1$$, because we supposed our theorem true for the aggregate $$E_{\nu-1}$$, with the cardinal number $$\nu$$,

(b) $$E'$$ consists of the single element $$e_\nu$$, then $$\overline\overline{E'}=1$$.

(c) $$E'$$ consists of $$e_\nu$$ and an aggregate $$E$$, so that $$E' = (E, e_\nu)$$. $$E''$$ is a part of $$E_{\nu-1}$$ and has therefore by supposition as cardinal number one of the numbers $$1, 2, 3,. . ., \nu-1$$. But now $$\overline\overline{E'} = \overline\overline{E''}+1$$, and thus the cardinal number of $$E'$$ is one of the numbers $$2, 3, ..., \nu$$.

Proof of A.—Every one of the aggregates which we have denoted by $$E_\nu$$ has the property of not being equivalent to any of its parts. For if we suppose that this is so as far as a certain $$\nu$$, it follows from the theorem D that it is so for the immediately following number $$\nu+1$$. For $$\nu=1$$, we recognize at once that the aggregate $$E_1 = (e_0, e_1)$$ is not equivalent to any of its parts, which are here $$(e_0)$$ and $$(e_1)$$. Consider, now, any two numbers $$\mu$$ and $$\nu$$ of the series $$1, 2, 3, ...$$; then, if $$\mu$$ is the earlier and $$\nu$$ the later, $$E_{\mu-1}$$ is a part of $$E_{\nu-1}$$. Thus $$E_{\mu-1}$$ and