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number of $N_{1}$ is equal to one of the preceding numbers $1,2,3,...,\nu -1$.

Proof of D. Suppose that the aggregate $(M,e)$ is equivalent to one of its parts which we will call $N$. Then two cases, both of which lead to a contradiction, are to be distinguished:

(a) The aggregate $N$ contains $e$ as element; let $N=(M_{1},e)$; then $M_{1}$ is a part of $M$ because $N$ is a part of $(M,e)$. As we saw in § 1, the law of correspondence of the two equivalent aggregates $(M,e)$ and $(M_{1},e)$ can be so modified that the element $e$ of the one corresponds to the same element $e$ of the other; by that, then, $M$ and $M_{1}$ are referred reciprocally and univocally to one another. But this contradicts the supposition that $M$ is not equivalent to its part $M_{1}$.

(b) The part $N$ of $(M,e)$ does not contain $e$ as element, so that $N$ is either $M$ or a part of $M$. In the law of correspondence between $(M,e)$ and $N$, which lies at the basis of our supposition, to the element $e$ of the former let the element $f$ of the latter correspond. Let $N=(M_{1},f)$; then the aggregate $M$ is put in a reciprocally univocal relation with $M_{1}$. But $M_{1}$ is a part of $N$ and hence of $M$. So here too $M$ would be equivalent to one of its parts, and this is contrary to the supposition.

Proof of E.—We will suppose the correctness of the theorem up to a certain $\nu$ and then conclude its validity for the number $\nu +1$ which immediately follows, in the following manner:—We start from the aggregate $E_{\nu }=(e_{0},e_{1},...,e_{\nu })$ as an aggregate