The cell form is associative and its radical is a submodule

The submodules $A^{⊵\lambda }$ and $A^{⊳\lambda }$ are two-sided ideals of $A$, and the following two congruence rules hold modulo $A^{⊳\lambda }$: for all $a\in A$ and $u,v\in T(\lambda )$, $c_{us}^{\lambda }a\equiv {\sum }_{w\in T(\lambda )}{r}_w{c}_{uw}^{\lambda },a{c}_{tv}^{\lambda }\equiv {\sum }_{w\in T(\lambda )}{s}_w{c}_{wv}^{\lambda },$ where the $r_w$ do not depend on $u$ and the $s_w$ do not depend on $v$. Indeed, the first is the cellular rule [F1] applied to the fixed label $\lambda$, the second-slot index $s$ and the element $a$, which gives coefficients independent of the free first index $u$; right multiplication by $a$ therefore preserves each $A^{⊵\mu }$ and $A^{⊳\mu }$, so $A^{⊵\lambda }$ and (as its span over all $\mu ⊳\lambda$) $A^{⊳\lambda }$ are right ideals, while applying the anti-automorphism $i$ to the first rule — it swaps the two indices, fixes $R$ and preserves both spans — gives the second rule and shows that they are also left ideals.

Existence and uniqueness of the form: for all $s,t\in T(\lambda )$ there is a unique $r=r(s,t)\in R$ with $c_{us}^{\lambda }{c}_{tv}^{\lambda }\equiv r{c}_{uv}^{\lambda }(\mathrm{m}\mathrm{o}\mathrm{d}{A}^{⊳\lambda })$ for all $u,v$, and $r$ depends only on $s,t$. Since $A^{⊳\lambda }$ is a two-sided ideal by step 1.1, the quotient ring $A\mathrm{/}{A}^{⊳\lambda }$ contains $Q:=A^{⊵\lambda }\mathrm{/}{A}^{⊳\lambda }$ with $R$-basis the classes ${\bar{c}}_{uv}:=c_{uv}^{\lambda }+A^{⊳\lambda }$. Applying the two rules of step 1.1 to the product $c_{us}^{\lambda }{c}_{tv}^{\lambda }$ — first with $a:=c_{tv}^{\lambda }$, then with $a:=c_{us}^{\lambda }$ — gives ${\bar{c}}_{us}{\bar{c}}_{tv}={\sum }_w{\rho }_w{\bar{c}}_{uw}={\sum }_x{\sigma }_x{\bar{c}}_{xv},$ where ${\rho }_w$ does not depend on $u$ and ${\sigma }_x$ does not depend on $v$. Comparing the coefficients of ${\bar{c}}_{ab}$ in the two bases: the left expansion contributes ${\rho }_b$ when $a=u$ and $0$ otherwise, the right expansion ${\sigma }_a$ when $b=v$ and $0$ otherwise; hence ${\sigma }_a=0$ for $a\mathrm{\ne }u$ and ${\rho }_b=0$ for $b\mathrm{\ne }v$, leaving ${\bar{c}}_{us}{\bar{c}}_{tv}={\sigma }_u{\bar{c}}_{uv}={\rho }_v{\bar{c}}_{uv}$. The common value ${\sigma }_u={\rho }_v$ is independent of $v$ (being ${\sigma }_u$) and independent of $u$ (being ${\rho }_v$), so it depends only on $s,t$; call it $⟨c_s^{\lambda },c_t^{\lambda }{⟩}_{\lambda }$. It is determined by the displayed congruence, because the coefficient of ${\bar{c}}_{uv}$ in the indicated free basis is unique, including over the zero ring. Bilinear extension gives the unique $R$-bilinear form with the stated property.

Symmetry: for all $s,t,u,v$, using that $i$ is an anti-automorphism with $i(c_{ab}^{\lambda })=c_{ba}^{\lambda }$, fixes $R$ and preserves $A^{⊳\lambda }$, one has $⟨c_s^{\lambda },c_t^{\lambda }{⟩}_{\lambda }{\bar{c}}_{uv}\equiv c_{us}^{\lambda }{c}_{tv}^{\lambda }=(c_{vt}^{\lambda }{c}_{su}^{\lambda }{)}^i\equiv (⟨c_t^{\lambda },c_s^{\lambda }{⟩}_{\lambda }{c}_{vu}^{\lambda }{)}^i=⟨c_t^{\lambda },c_s^{\lambda }{⟩}_{\lambda }{\bar{c}}_{uv},$ so $⟨c_s^{\lambda },c_t^{\lambda }{⟩}_{\lambda }=⟨c_t^{\lambda },c_s^{\lambda }{⟩}_{\lambda }$, and by bilinearity $⟨x,y{⟩}_{\lambda }=⟨y,x{⟩}_{\lambda }$ for all $x,y\in C^{\lambda }$.

Associativity: fix $s,t\in T(\lambda )$ and $a\in A$, and write $c_s^{\lambda }a={\sum }_w{r}_w{c}_w^{\lambda }$ in $C^{\lambda }$ — the coefficients $r_w$ are those of step 1.1. Then, by step 2.1 applied first to $c_s^{\lambda }a$ and bilinearity of the form, $⟨c_s^{\lambda }a,c_t^{\lambda }{⟩}_{\lambda }{\bar{c}}_{uv}={\sum }_w{r}_w{\bar{c}}_{uw}{\bar{c}}_{tv}=({\bar{c}}_{us}\bar{a})\cdot {\bar{c}}_{tv}={\bar{c}}_{us}\cdot \left(\bar{a}{\bar{c}}_{tv}\right)\mathrm{.}$ The second rule of step 1.1 writes $a{c}_{tv}^{\lambda }\equiv {\sum }_x{s}_x{c}_{xv}^{\lambda }(\mathrm{m}\mathrm{o}\mathrm{d}{A}^{⊳\lambda })$ with coefficients $s_x$ characterised by $c_t^{\lambda }i(a)={\sum }_x{s}_x{c}_x^{\lambda }$ in $C^{\lambda }$; hence ${\bar{c}}_{us}\cdot \left(\bar{a}{\bar{c}}_{tv}\right)={\sum }_x{s}_x{\bar{c}}_{us}{\bar{c}}_{xv}={\sum }_x{s}_x⟨c_s^{\lambda },c_x^{\lambda }{⟩}_{\lambda }{\bar{c}}_{uv}=⟨c_s^{\lambda },c_t^{\lambda }i(a){⟩}_{\lambda }{\bar{c}}_{uv}=⟨c_s^{\lambda },a\cdot c_t^{\lambda }{⟩}_{\lambda }{\bar{c}}_{uv}\mathrm{.}$ Cancelling the basis element ${\bar{c}}_{uv}$ gives $⟨c_s^{\lambda }a,c_t^{\lambda }{⟩}_{\lambda }=⟨c_s^{\lambda },i(a)c_t^{\lambda }{⟩}_{\lambda }$; by bilinearity this is $⟨xa,y{⟩}_{\lambda }=⟨x,i(a)y{⟩}_{\lambda }$ for all $x,y\in C^{\lambda }$. Replacing $a$ by $i(a)$ and using $i^2=\mathrm{i}\mathrm{d}$ gives the equivalent left-action form $⟨xi(a),y{⟩}_{\lambda }=⟨x,i(a)\cdot y{⟩}_{\lambda }$, where $i(a)\cdot y=yi(i(a))=ya$ is the left action of $i(a)$: this is the dual statement.

The adjointness identity $x{c}_{uv}^{\lambda }=⟨x,c_u^{\lambda }{⟩}_{\lambda }{c}_v^{\lambda }$ holds for all $x\in C^{\lambda }$ and $u,v\in T(\lambda )$: it suffices to treat $x=c_s^{\lambda }$, and there step 2.1 with $(u,s,t,v):=(s,s,u,v)$ gives $c_{ss}^{\lambda }{c}_{uv}^{\lambda }\equiv ⟨c_s^{\lambda },c_u^{\lambda }{⟩}_{\lambda }{c}_{sv}^{\lambda }(\mathrm{m}\mathrm{o}\mathrm{d}{A}^{⊳\lambda })$; by the definition of the right action in [F1], the coefficients of $c_{sv}^{\lambda }$ on the right are exactly the coefficients of $c_v^{\lambda }$ in $c_s^{\lambda }\cdot c_{uv}^{\lambda }$, so $c_s^{\lambda }{c}_{uv}^{\lambda }=⟨c_s^{\lambda },c_u^{\lambda }{⟩}_{\lambda }{c}_v^{\lambda }$; bilinearity in $x$ extends this to all $x$.

The radical is a submodule and a kernel: for fixed $y\in C^{\lambda }$ the assignment $x\mapsto ⟨x,y{⟩}_{\lambda }$ is an $R$-linear map with kernel $\{x:⟨x,y{⟩}_{\lambda }=0\}$, so $\mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }={\bigcap }_y\ker (x\mapsto ⟨x,y{⟩}_{\lambda })$ is an $R$-submodule; and if $x\in \mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }$ then for every $a\in A$ and $y\in C^{\lambda }$ step 3.2 gives $⟨xa,y{⟩}_{\lambda }=⟨x,i(a)y{⟩}_{\lambda }=0$, so $xa\in \mathrm{r}\mathrm{a}\mathrm{d}{C}^{\lambda }$: the radical is an $A$-submodule, namely the kernel of the $A$-module map $x\mapsto ⟨x,-{⟩}_{\lambda }$. This completes the proof of (i)-(iv).