with two $d$s in different rows. If $T$ is $d$-bad, then we can express ${\psi_{d,1}}\circ{\hat\Theta_{T}}$ as a linear combination of semistandard homomorphisms using Lemma \[lemma5\] together with Lemma \[newsemilem\]. For example, if $$\begin{aligned}
T&=\young(11111122248{10},369{11}{12},57)
\\
\intertext{then ${\psi_{6,1}}\circ{\hat\Theta_{T}}={\hat\Theta_{T'}}$, where}
T'&=\young(11111122248{10},369{11}{12},56)\end{aligned}$$ and we can semistandardise this using Lemma \[newsemilem\], taking $$A=\{3\},\qquad B=\{5,6,7,9,11,12\},\qquad C=\emptyset$$ to express ${\hat\Theta_{T'}}$ as a sum of fourteen semistandard homomorphisms.

Doing this for each $d$-bad tableau $T$, we find that ${\psi_{d,1}}\circ{\hat\Theta_{T}}$ is a sum of homomorphisms labelled by semistandard tableaux with the same first row as $T$; furthermore, at least one of these tableaux will have two $d$s in the second row. Moreover, each $d$-bad tableau will yield a tableau of this kind which does not occur for any other $d$-bad tableau $T'$. To see this, suppose first of all that $d,d+1$ occur in the second column of $T$. Then there is no other $d$-bad tableau with the same first row as $T$, so any tableau occurring in ${\psi_{d,1}}\circ{\hat\Theta_{T}}$ with two $d$s in the second row can only possibly occur in ${\psi_{d,1}}\circ{\hat\Theta_{T}}$. Alternatively, if $d,d+1$ occur in the first column of $T$, then $T$ has the form $${\text{\footnotesize$\gyoungx(1.2,;1_5{{\begin{tikzpicture}[baseline=0cm]\draw[thick,densely dotted](0,.13)--++(5*.25,0);\end{tikzpicture}}};1;2;2;2;{x_1}_{1.5}{{\begin{tikzpicture}[baseline=0cm]\draw[thick,densely dotted](0,.13)--++(1.5*.25,0);\end{tikzpicture}}};{x_s},;d;{z_1};{z_2};{{\begin{tikzpicture}[baseline=0cm]\draw[thick,densely dotted](0,.13)--++(1*.25,0);\end{tikzpicture}}};{z_t},;{d\!\!+\!\!1};k)$}}.$$ There are $v-2$ other $d$-bad tableaux with the same first row as $T$, and they all also have the same $(2,2)$-entry as $T$. Hence when we apply Lemma \[lemma5\] and Lemma \[newsemilem\] (or equivalently 