[proofplan] The result is precisely the row-insertion characterization of Knuth equivalence. We apply the [Knuth equivalence theorem](/theorems/8432) for Schensted row insertion, after checking that its ordered alphabet, semistandard tableau, insertion, and congruence conventions match the conventions in the statement. [/proofplan] [step:Apply the Knuth equivalence theorem with the present conventions] The ordered alphabet in the statement is the totally ordered set $(A,\le)$, the word monoid is $A^*$ with identity word $\epsilon$, and the set $\operatorname{SSYT}(A)$ consists of finite tableaux with weakly increasing rows and strictly increasing columns. The operation $T\leftarrow a$ is Schensted row insertion with the leftmost-strictly-larger bumping convention, and $P(w)$ is obtained by iterating this insertion from the empty tableau $\varnothing$. These are exactly the hypotheses and conventions of [citetheorem:8432]. Therefore, for every $u,v\in A^*$, \begin{align*} u\sim_K v \iff P(u)=P(v). \end{align*} [guided] We do not need to reprove the local bumping-path and row-word machinery here. The theorem [citetheorem:8432] is stated for a totally ordered alphabet $(A,\le)$, the free monoid $A^*$ of finite words, semistandard Young tableaux with weak rows and strict columns, and Schensted row insertion using the same leftmost-entry-strictly-larger-than-the-inserted-letter convention. The present statement defines the same objects: $A^*$ is the word monoid, $\operatorname{SSYT}(A)$ is the tableau set, $T\leftarrow a$ is the same row insertion operation, and $P(w)$ is the tableau obtained by successively inserting the letters of $w$ starting from the empty tableau. The congruence $\sim_K$ is also generated by the same two elementary Knuth relations, \begin{align*} xzy \sim_K zxy \quad \text{whenever } x\le y<z, \end{align*} and \begin{align*} yxz \sim_K yzx \quad \text{whenever } x<y\le z. \end{align*} Thus all hypotheses of [citetheorem:8432] are satisfied, and its conclusion gives exactly \begin{align*} u\sim_K v \iff P(u)=P(v) \end{align*} for all words $u,v\in A^*$. [/guided] [/step]