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For nonnegative functions on a finite Boolean lattice, the condition $\alpha (A)\beta (B)\le \gamma (A\cup B)\delta (A\cap B)$ for all pairs implies $(\sum \alpha )(\sum \beta )\le (\sum \gamma )(\sum \delta )$. Coordinate elimination and the two-point four-functions inequality prove it by induction. Indicators give inequalities for the union and intersection of set families. The Boolean-lattice statement appears in Theorem 3.1 of Nicholas Ruozzi's paper.

The four functions theorem concerns nonnegative functions $\alpha ,\beta ,\gamma ,\delta$ on the Boolean lattice. Its pointwise hypothesis and conclusion areWe prove it by eliminating coordinates.

The elementary step is the two-point four-functions inequality. Suppose $a_i{b}_j\le c_{\max (i,j)}{d}_{\min (i,j)}$ for $i,j\in \{0,1\}$. Put $x=a_0{b}_1$, $y=a_1{b}_0$, $M=c_1{d}_0$, $R=c_0{d}_1$. Then $x,y\le M$ and, using the two diagonal bounds,If $M>0$, $(M-x)(M-y)\ge 0$ yields $x+y\le M+xy/M\le M+R$. If $M=0$, both cross terms vanish and the same conclusion holds. Adding the diagonal bounds gives

For subsets $S,T\subseteq [n-1]$, take $a_i=\alpha (S\cup E_i)$, $b_j=\beta (T\cup E_j)$, $c_i=\gamma ((S\cup T)\cup E_i)$ and $d_i=\delta ((S\cap T)\cup E_i)$, where $E_0=\varnothing$, $E_1=\{n\}$. The original hypothesis supplies precisely the four inequalities of the elementary step. Therefore the marginal functionsand similarly ${\beta }^{\prime },{\gamma }^{\prime },{\delta }^{\prime }$, satisfy the same pointwise hypothesis on $[n-1]$. Induction, starting with the one-element Boolean lattice on the empty ground set, proves the four functions theorem. This proof includes zero-valued functions without division by zero.

For the Harris-Kleitman inequality, let $\mathcal{F},\mathcal{G}$ be increasing set families, meaning they contain every superset of each member. In the theorem take $\alpha =1_{\mathcal{F}}$, $\beta =1_{\mathcal{G}}$, $\gamma =1_{\mathcal{F}\cap \mathcal{G}}$ and $\delta =1$. If $A\in \mathcal{F}$ and $B\in \mathcal{G}$, their union belongs to both increasing families, so the hypothesis holds. Thus, for the uniform probability measure $\mu$ on subsets,Mapping every subset to its complement proves the same Harris-Kleitman inequality for two decreasing families. An increasing and a decreasing family have the reverse inequality: apply positive correlation to the increasing family and the complement, as a family, of the decreasing one. More generally, increasing nonnegative functions obey the corresponding positive-correlation bound by the same four-functions substitution, since both function values at the union dominate the original values. For an independent-coordinate product measure, multiply all four functions by its weight $w$; the identity $w(A)w(B)=w(A\cup B)w(A\cap B)$ preserves the hypothesis and proves the weighted Harris inequality too.

Finally, replace each intersecting family ${\mathcal{A}}_i$ by its upward closure of a set family ${\mathcal{U}}_i$. It is still intersecting: two supersets of intersecting members have nonempty intersection. It contains at most one of each pair $S,S^c$, and hence $|{\mathcal{U}}_i|\le 2^{n-1}$. Let ${\mathcal{D}}_i$ be the complement of ${\mathcal{U}}_i$ as a family. These families are decreasing and have measure at least one half. Their intersections are decreasing too, so repeated Harris-Kleitman inequality givesTaking complements and using ${\mathcal{A}}_i\subseteq {\mathcal{U}}_i$ proves the union bound for intersecting families:For $k\le n$, equality is attained by the families of sets containing coordinate $i$, $1\le i\le k$. For $k>n$, the displayed real bound implies the integer bound $2^n-1$, which is attained by using the $n$ coordinate families and repeating families as necessary.

The Ahlswede–Daykin inequality, also called the four functions theorem, concerns nonnegative functions $\alpha ,\beta ,\gamma ,\delta$ on $\mathcal{P}([n])$. IfthenProve it by induction on $n$. The case $n=0$ is the single assumed inequality. For the induction step, sum each function over the last coordinate to obtain ${\alpha }^{\prime },{\beta }^{\prime },{\gamma }^{\prime },{\delta }^{\prime }$ on $\mathcal{P}([n-1])$. For fixed $A,B$ in this smaller cube, set $a_i=\alpha (A\cup i\{n\})$, $b_i=\beta (B\cup i\{n\})$, $c_i=\gamma ((A\cup B)\cup i\{n\})$, and $d_i=\delta ((A\cap B)\cup i\{n\})$, where $i\{n\}$ is empty for $i=0$ and $\{n\}$ for $i=1$.

The hypotheses give $a_0{b}_0\le c_0{d}_0$, $a_1{b}_1\le c_1{d}_1$ and $a_0{b}_1,a_1{b}_0\le c_1{d}_0$. The two-point four-functions inequality shows that these imply $(a_0+a_1)(b_0+b_1)\le (c_0+c_1)(d_0+d_1)$. Here is its key algebra: put $x=a_0{b}_1$, $y=a_1{b}_0$, $M=c_1{d}_0$, $N=c_0{d}_1$. Then $x,y\le M$ and $xy\le MN$. If $M>0$, $(M-x)(M-y)\ge 0$ gives $x+y\le M+xy/M\le M+N$; if $M=0$, both cross terms vanish. Adding the two diagonal bounds proves the claim.

Thus the primed functions satisfy the same hypothesis, and the induction hypothesis applies. Their totals equal the original totals, completing the proof.