Removing a corner changes hooks in its row and column

For these coordinate comparisons, extend row lengths by zero beyond the last nonempty row. The row lengths of $\mu$ are ${\mu }_a={\lambda }_a-1=b-1$ and ${\mu }_i={\lambda }_i$ for $i\mathrm{\ne }a$: deleting the row-end box of row $a$ shortens exactly that row, and the result is a partition by [L2]. The column heights are ${\mu }_b^{\mathrm{\prime }}={\lambda }_b^{\mathrm{\prime }}-1=a-1$ and ${\mu }_j^{\mathrm{\prime }}={\lambda }_j^{\mathrm{\prime }}$ for $j\mathrm{\ne }b$: column $b$ loses exactly its bottom box, since row $a$ is the last row of length at least $b$ (rows below row $a$ have length $<b$ by [L3] and removability), while a column $j\mathrm{\ne }b$ either still meets row $a$ (if $j<b$, when row $a$ has length $b-1\ge j$) or never met row $a$ (if $j>b$, when row $a$ has length $b<j$), so its height is unchanged.

Every $y\in R_x$ satisfies $h_{\lambda }(y)\ge 2$: a box of row $a$ at column $j<b$ is not the end of its row, and a box $(i,b)$ with $i<a$ has the box $(i+1,b)$ of $[\lambda ]$ directly below it, since ${\lambda }_{i+1}\ge {\lambda }_a=b$ for $i+1\le a$; in both cases $y$ is not removable, so $h_{\lambda }(y)\mathrm{\ne }1$ and, being positive, $h_{\lambda }(y)\ge 2$.

For $y=(i,j)\in [\mu ]$ with $i\mathrm{\ne }a$ and $j\mathrm{\ne }b$, both summands of $h(y)={\nu }_i-j+{\nu }_j^{\mathrm{\prime }}-i+1$ are the same for $\nu =\lambda$ and for $\nu =\mu$, so $h_{\mu }(y)=h_{\lambda }(y)$.

For $y=(a,j)\in [\mu ]$ with $j<b$ one has $h_{\mu }(y)={\mu }_a-j+{\mu }_j^{\mathrm{\prime }}-a+1=({\lambda }_a-1)-j+{\lambda }_j^{\mathrm{\prime }}-a+1=h_{\lambda }(y)-1$, because $j\mathrm{\ne }b$ leaves the column height unchanged.

For $y=(i,b)\in [\mu ]$ with $i<a$ one has $h_{\mu }(y)={\mu }_i-b+{\mu }_b^{\mathrm{\prime }}-i+1={\lambda }_i-b+({\lambda }_b^{\mathrm{\prime }}-1)-i+1=h_{\lambda }(y)-1$, because $i\mathrm{\ne }a$ leaves the row length unchanged.

The multiset of hook factors: $h_{\lambda }(x)=1$, so $P(\lambda )={\prod }_{y\in [\mu ]}{h}_{\lambda }(y)\cdot h_{\lambda }(x)={\prod }_{y\in [\mu ]}{h}_{\lambda }(y)$, while $P(\mu )={\prod }_{y\in [\mu ]}{h}_{\mu }(y)=({\prod }_{y\in [\mu ]\setminus R_x}{h}_{\lambda }(y))({\prod }_{y\in R_x}(h_{\lambda }(y)-1))$. Dividing the two finite products, all factors with $y\mathrm{\notin }{R}_x$ cancel and the factors with $y\in R_x$ contribute $h_{\lambda }(y)\mathrm{/}(h_{\lambda }(y)-1)$; the division is legitimate because $h_{\lambda }(y)-1\ge 1$ on $R_x$ by step 1.2.