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$E_{\nu -1}$ are not equivalent, and accordingly their cardinal numbers $\mu ={\overline {\overline {E}}}_{\mu -1}$ and $\nu ={\overline {\overline {E}}}_{\nu -1}$ are not equal.

Proof of B.—If of the two finite cardinal numbers $\mu$ and $\nu$ the first is the earlier and the second the later, then $\mu <\nu$. For consider the two aggregates $M=E_{\mu -1}$ and $N=E_{\nu -1}$; for them each of the two conditions in § 2 for ${\overline {\overline {M}}}<{\overline {\overline {N}}}$ is fulfilled. The condition (a) is fulfilled because, by theorem E, a part of $M=E_{\mu -1}$ can only have one of the cardinal numbers $1,2,3,...,\mu -1$, and therefore, by theorem A, cannot be equivalent to the aggregate $N=E_{\mu -1}$. The condition (b) is fulfilled because $M$ itself is a part of $N$.

Proof of C.— Let ${\mathfrak {a}}$ be a cardinal number which is less than $\nu +1$. Because of the condition (b) of §2, there is a part of $E_{\nu }$ with the cardinal number ${\mathfrak {a}}$. By theorem E, a part of $E_{\nu }$ can only have one of the cardinal numbers $1,2,3,...,\nu$. Thus ${\mathfrak {a}}$ is equal to one of the cardinal numbers $1,2,3,...,\nu$. By theorem B, none of these is greater than $\nu$. Consequently there is no cardinal number ${\mathfrak {a}}$ which is less than $\nu +1$ and greater than $\nu$.

Of importance for what follows is the following theorem:

F. If $K$ is any aggregate of different finite cardinal numbers, there is one, $\kappa _{1}$, amongst them which is smaller than the rest, and therefore the smallest of all.

[492] Proof—The aggregate $K$ either contains