Let $X_1,X_2,\ldots,X_n $ be indeterminates. Denote by $S$ the set of all linear inequalities of the form $X_{i_1}+X_{i_2}+\ldots+X_{i_k} \geq k,$ with $k \in \{ 1,2,\ldots,n \}$ and $1 \leq i_1< i_2 < \ldots < i_k \leq n $.

Consider the following illustrative example. We are given constraints $$ D=\{X_1+X_2+X_3\geq 3, X_1+X_3+X_4 \geq 3, X_1+X_2 + X_4 \geq 3, X_2+X_3+X_4 \geq 3\} \subset S. $$ We can infer that the inequality $$ X_1+X_2+X_3+X_4 \geq 4, $$ another member of $S$, also holds via a "complicated" nonintegral linear combination: $$ \frac{1}{3}(X_1+X_2+X_3)+\frac{1}{3}(X_1+X_3+X_4)+\frac{1}{3}(X_2+X_3+X_4)+\frac{1}{3}(X_1+X_3+X_4) \geq 4 $$

In general, it is conceivable that we can be given some set of constraints $D$ and can only infer an inequality in $S$ via a very complicated and nonobvious $\mathbb{R}_+$-linear combination. Indeed, deducing an inequality is much like solving an LP. I am interested in a simpler algorithm for deducing all inequalities in $S$ from those given in $D$.

One idea I had for such an algorithm is as follows. Notice that $$ X_i+X_j+X_k \geq 3 \implies (X_i \geq 1) \vee (X_j \geq 1) \vee (X_k \geq 1). $$ For instance, in our case, we know that $$ X_1+X_2+X_3 \geq 3 \implies (X_1 \geq 1) \vee (X_2 \geq 1) \vee (X_3 \geq 1) $$ Therefore we know that the following logical expression is true $$ ((X_1 \geq 1) \vee (X_2 \geq 1) \vee (X_3 \geq 1)) \wedge (X_2+X_3+X_4 \geq 3) \wedge (X_1+X_3+X_4 \geq 3) \wedge (X_1+X_3+X_4 \geq 3). $$ Expanding, and using $$ (X_a \geq 1) \wedge (X_b+X_c+X_d \geq 3) \implies (X_a+X_b+X_c+X_d \geq 4), $$ which amounts to taking an ordinary (having coefficients $+1$) sum, this implies that $$ X_1+X_2+X_3+X_4 \geq 4. $$ More generally, we can infer from $$ X_{i_1}+X_{i_2}+\ldots+X_{i_k} \geq k $$ that $$ (X_{i_2}+\ldots+X_{i_k} \geq k-1) \vee (X_{i_1}+X_{i_3}+\ldots+X_{i_k} \geq k-1) \vee \ldots $$ and we can continue on inductively. One would like to show that no matter which selection of these inequalities we make, assuming everything is consistent, we can sum them to obtain any inequality that can be deduced via a general linear combination. So the question is

${\bf Question}$: Can every inequality that holds be deduced by adding such "simple" inequalities obtained via the procedure above?

${\bf Remark}$: there are many ways to interpret this question, which leads me to think that something of the sort ought to be known. E.g, in terms of average, in terms of the $n$-hypercube, and in terms of (0,1) matrices.

${\bf Clarification}$: any of the inequalities from $D$ can be used in the disjoint union.