Graph theory - Not sure what "all but at most" means 
Given $d\in \text{[natural numbers]}$ and $G$ a bipartite graph with vertex classes $A$ and $B$ so that $|N(S)|\geq |S|-d$ for all $S\subseteq A$, deduce that $G$ has a matching covering all but at most $d$ vertices in $A$.

Not sure what "...all but at most $d$ vertices ..." means? In a sense it implies $|A|\leq d$ if a matching covers all of $A$ and $|\text{matching}|\leq d$, but that sounds a bit absurd.
Could someone kindly explain this?
 A: More a question of English than maths.  I expect that it could have been written more clearly but it seems to be based on more familiar "all but" phrases.  
For illustration, let's suppose that A has 10 vertices. 
"All but one vertex" would mean 9.  
"All but two vertices" would mean 8.  
"All but d vertices" would mean 10 - d.  
"All but at most one vertex" would mean 9 or 10.  
"All but at most two vertices" would mean 8, 9, or 10.  
"All but at most d vertices" would mean from 10 - d to 10.  
A: The conclusion can be rephrased as "$G$ has a matching of size $|A|-d$" or as "$G$ contains $|A|-d$ independent edges" or as "$G$ contains at least $|A|-d$ independent edges.  
The assertion can be proved by adding $d$ new vertices to $B$, joining these vertices to each vertex in $A$, and applying Hall's theorem. Hall's theorem ensures that this augmented graph has $|A|$ independent edges, and since up to $d$ of these independent edges could be incident to the new vertices, the original graph is guaranteed to contain at least $|A|-d$ independent edges.  Thus, up to $d$ vertices of $A$ might not be incident to these independent edges.
In other words, there is a matching in $G$ which covers all vertices of $A$ except for a subset $A' \subseteq A$ of size at most $d$.  The matching would therefore cover all vertices in $A-A'$, which has size at least $|A|-d$ (because $|A'| \le d$ implies $|A-A'| \ge |A|-d$).       
