time we write $S^\lambda \in H^*(Gr(a,b))$ (or otherwise assume that $\lambda$ indexes such a class), then we implicitly assume $$\lambda = \mathbf{0} \leq \lambda_1  \leq \cdots \leq \lambda_a \leq
\mathbf{b-a} \in \Lambda(a,b-a).$$

\[def:hornineq\] Let $\lambda^1, \ldots, \lambda^s \in \Lambda(r,n-r)$ Suppose that for every $d$, $1 \leq d \leq r$, and every non-zero product $S^{\mu^1} \cdots S^{\mu^s} \in H^*(Gr(d,r))$, the inequality $$\label{eqn:hornineq}
\sum_{i=1}^s \sum_{k=1}^d \lambda^i_{\mu^i_k +k} \geq (s-1)d(n-r)$$ holds. In this case we say that the [**Horn inequalities**]{} hold for $\lambda^1, \ldots, \lambda^s$.

The product $S^{\lambda^1} \cdots S^{\lambda^s} \in H^*(Gr(r,n))$ is non-zero if and only if the Horn inequalities hold for $\lambda^1, \ldots, \lambda^s$.

The standard form of Horn’s conjecture assumes that the product of $S^{\lambda^1} \cdots S^{\lambda^s}$ is of top degree, whereas this formulation (which appears in [@B]) does not. To accommodate this, definition \[def:hornineq\] allows an inequality for all non-zero products in $H^*(Gr(d,r))$ rather than those just those of top degree. This generalisation is easily shown to imply the standard statement.

Definition \[def:hornineq\] differs slightly from the usual definition of the Horn inequalities in another small way. Normally one does not include an inequality for the case $d=r$; one simply uses $d$ for which $1 \leq d \leq r-1$. 