As usual, pick any intertwinor and write
Now define the set to consist of those semistandard generalized tableaux so that for some appearing in this sum with a nonzero coefficient. This is called the “lower order ideal” generated by the in the sum. We will prove our assertion by induction on the size of this order ideal.
If is empty, then must be the zero map. Indeed, our lemmas showed that if is not the zero map, then at least one semistandard shows up in the above sum, and this would itself belong to . And of course the zero map is contained in any span.
Now, if is not empty, then there is at least some semistandard with in the sum. Our lemmas even show that we can pick one so that is maximal among all the tableaux in the sum. Let’s do that and define a new intertwinor:
I say that is with removed.
Every appearing in has , since if is semistandard then is the largest column equivalence class in . Thus must be a subset of since we can’t be introducing any new nonzero coefficients.
Our lemmas show that if , then must appear with the same coefficient in both and . That is, they must be cancelled off by the subtraction. Since is maximal there’s nothing above it that might keep it inside the ideal, and so .
So by induction we conclude that is contained within the span of the generated by semistandard tableaux, and thus must be as well.