TITLE: Box removing operators on partitions
QUESTION [11 upvotes]: Consider the box removing operators $d_i$ for $i\geq 1$ defined as follows.
Let $\lambda$ be a partitions. Then $d_i(\lambda)$ is the partition obtained by removing a box from a part of length $i$ in $\lambda$ so as to obtain another partition. (Here removing a box is an operation performed on the Ferrers diagram of $\lambda$.)
If there is no part of length $i$ in $\lambda$, then $d_i(\lambda)=0$.
(As pointed out by Darij in the comments below, $0$ is not to be confused with the empty partition $\varnothing$.)
It is easy to see that the following equalities are satisfied.
$$
d_id_j=d_jd_i \text{ if } \lvert i-j \rvert\geq 2$$
$$d_id_{i+1}d_i=d_id_id_{i+1}$$
$$d_id_{i+1}d_{i+1}=d_{i+1}d_id_{i+1}$$ 
Now consider words in the $d_i$. Two words $w_1$ and $w_2$ are called equivalent if $w_1(\lambda)=w_2(\lambda)$ for all partitions $\lambda$.
Is it true that if $w_1, w_2$ are equivalent then one can be obtained from the other by using the 3 relations mentioned above?

REPLY [4 votes]: These are the Coxeter relations, and I believe equivalence comes from a general theorem: Tit's Theorem. For example, if you are willing to accept the Edelemann-Greene correspondence between Young Tableau and reduced words of permutations, then it would be equivalent to show that one can move freely between two reduced words using the above relations (now correspondingly, $d_i$ act on permutations via adjacent transpositions: $d_i(i,i+1)=(i+1,i)$. You can find a proof of reduced word equivalence here, page 4, Theorem 1.1.2. Essentially, the proof for tableaux should be of the same flavor: show that you can pass from any word to a carefully chosen canonical word.