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Source. Ford–Fulkerson (1958), printed p. 83, the deduction following Theorem 2; the footnote attached to the converse credits that short proof to O. Gross (published scan).

The inequalities of Theorem 2 imply those of Theorem 1 for each family. If $S_j=T_j$ at every index, the two sets of tests are equivalent.

Proof. For the first implication fix $X$ and take $Y=\varnothing$ in Theorem 2. This gives

$$|X|\le n-a+\alpha(I(X)).$$

Taking $Y=[n]$ instead and subtracting $n$ gives

$$|X|\le-a+\alpha(I(X)\cup J([n])) +\beta(I(X)\cap J([n]))\le\beta(I(X)).$$

The last inequality uses $\alpha(I(X)\cup J([n]))\le a$ and nonnegativity of the $\beta_i$. These are precisely the two tests for $\mathcal S$. Interchanging the families gives the tests for $\mathcal T$.

Now assume the families are identical and their Theorem 1 tests hold. For arbitrary $X,Y$, apply its lower-bound test to $X\cup Y$ and its upper-bound test to $X\cap Y$. Since

$$I(X\cup Y)=I(X)\cup I(Y),\qquad I(X\cap Y)\subseteq I(X)\cap I(Y),$$

we obtain

$$\begin{aligned} |X\cup Y|&\le n-a+\alpha(I(X)\cup I(Y)),\\ |X\cap Y|&\le\beta(I(X\cap Y)) \le\beta(I(X)\cap I(Y)). \end{aligned}$$

Adding and using $|X|+|Y|=|X\cup Y|+|X\cap Y|$ gives every Theorem 2 inequality. $\square$

Scope. The neighbor set of an intersection need only be contained in the intersection of the neighbor sets. No equality there is assumed. This algebraic deduction preserves the source's distinct explanation of the specialization; it is not a second proof of the external max-flow theorem.

Bears on. Comparing single-family quota tests with the stronger two-family system without duplicating their network proofs. No Erdős problem: the paper states no relation to a numbered Erdős problem.