= Solution
First prove preservation of the row and column inequalities for distinct ordered labels. If $b>c_l$, <row insertion> appends it to the first row, and the inequalities remain valid. Otherwise let $j$ be the first position with $c_j>b$. Replace $c_j$ by $b$ and insert the bumped $c_j$ in the second row. Its first larger entry is some $d_k$ with $k\le j$, because $d_j>c_j$. Replace $d_k$ by $c_j$ and put the bumped $d_k$ in the new third-row cell.
The first-larger-entry choices preserve strict row order. At column $k$, if $k<j$ the first-row entry is $c_k<c_j$, while if $k=j$ it is the new $b<c_j$. At column $j$ when $k<j$, the unchanged lower entry satisfies $d_j>c_j>b$. Every other old column keeps its previous inequalities. The new third-row entry is larger than the second-row first entry: for $k=1$ this is $d_1>c_j$, while for $k>1$ it follows from $d_k>d_1$. Thus the new shape $(l,l,1)$ is a <Young diagram> with strictly increasing rows and columns. This proves the insertion assertion for the ordered-label convention.
There is a numbering qualification in the literal standard convention of part (a). If $T$ uses exactly $1,\ldots,2l$, an arbitrary missing natural number $b$ need not be $2l+1$. For example, inserting six into rows $(1,2);(3,4)$ produces $(1,2,6);(3,4)$, which has the order inequalities but not the labels $1,\ldots,5$. Thus \b[standardness with consecutive labels requires $b=2l+1$, or increasing relabelling afterward]. The printed wording is retained in the private transcription; <row insertion> itself preserves the needed order structure.
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