Reversing a word transposes its insertion tableau

A finite real alphabet has a unique increasing enumeration. Its rank map preserves and reflects all inequalities, so the first-greater position, each carried label, and the final shape are unchanged under relabelling, by induction over the finite procedure; compressing the entry ranks gives the published standard tableau. Thus strict-row/strict-column arguments apply to the original real labels as well. (Base case and first-letter recursion.) $P(x_1)=[x_1]=x_1\to \mathrm{\varnothing }$: by [L2] with $T=\mathrm{\varnothing }$ one has $x_1\to \mathrm{\varnothing }=({\mathrm{\varnothing }}^{\mathrm{t}}\leftarrow x_1{)}^{\mathrm{t}}$, the insertion appends $x_1$ as the only box, and a one-box tableau equals its transpose; hence for $n=1$ the recursion holds.

(First-letter recursion, inductive step.) Let $n\ge 2$ and assume the recursion for words of length $n-1$. Then $P(x_1,\dots ,x_n)=P(x_1,\dots ,x_{n-1})\leftarrow x_n=(x_1\to P(x_2,\dots ,x_{n-1}))\leftarrow x_n=x_1\to (P(x_2,\dots ,x_{n-1})\leftarrow x_n)=x_1\to P(x_2,\dots ,x_n)$: the first and last equalities are [L1], the second is the induction hypothesis, and the third is [L3] applied to the standard tableau $T=P(x_2,\dots ,x_{n-1})$ with $x=x_1$, $y=x_n$, legitimate because the letters are pairwise distinct, so $x_1\mathrm{\notin }T$ and $x_1\mathrm{\ne }{x}_n$.

(Reversal, base cases.) For $n=0$ both sides are $\mathrm{\varnothing }$ and ${\mathrm{\varnothing }}^{\mathrm{t}}=\mathrm{\varnothing }$; for $n=1$ the one-box tableau equals its transpose.

(First-letter recursion, conclusion.) By steps 1.1 and 1.2 the recursion $P(x_1,\dots ,x_n)=x_1\to P(x_2,\dots ,x_n)$ holds for every $n\ge 1$ and every word of distinct letters.

(Reversal, inductive step.) Let $n\ge 2$ and assume $P(x_1,\dots ,x_k)=P(x_k,\dots ,x_1{)}^{\mathrm{t}}$ for all $k<n$. Then $P(x_n,\dots ,x_1{)}^{\mathrm{t}}=(x_n\to P(x_{n-1},\dots ,x_1){)}^{\mathrm{t}}=P(x_{n-1},\dots ,x_1{)}^{\mathrm{t}}\leftarrow x_n=P(x_1,\dots ,x_{n-1})\leftarrow x_n=P(x_1,\dots ,x_n)$: the first equality is the first-letter recursion of step 2.1 applied to the reversed word, the second is the definitional identity [L2], the third is the induction hypothesis, and the fourth is [L1].

(Conclusion.) By steps 1.3 and 3.1, $P(x_1,\dots ,x_n)=P(x_n,\dots ,x_1{)}^{\mathrm{t}}$ for every word of pairwise distinct real numbers; conversely the transpose relation for all words implies the first-letter recursion by reading the computation of step 3.1 backwards after replacing $(x_1,\dots ,x_n)$ by the reversed word, so the two displayed forms are equivalent.

(Recording tableau.) The statement makes no assertion about the recording tableau, and indeed the transpose relation is special to the insertion tableau: the recording tableau records the order in which boxes are added, and this order is not reversed by reversing the word (Schensted's note; see the example after Lemma 7). Nothing beyond the insertion tableau is claimed or used.