Four triplet fields strongly generate the complete singular span

Put $a_n=h_{1,2n+1}$ and $b_m=h_{1,-2m+1}$. For $X\in \{F,H,E\}$, let $A_X$ be its Virasoro state submodule. For each simple factor $L(c,a_n)$ of $S_0$ and quotient factor $L(c,b_m)$, the actual field product followed by quotient projection is a formal Virasoro intertwiner by [F2]. If $n\ge 1$, [F1] allows only $a_{n-1},a_n,a_{n+1}$. For $k\ge 0$, equality $h_{1,2k+1}=h_{1,-2m+1}$ would imply $(1-(2k+1)p{)}^2=(1+(2m-1)p{)}^2\mathrm{.}$ Taking the two possible signs gives either $k=-m$, impossible for $k\ge 0,m\ge 1$, or $p(k-m+1)=1$, impossible for integers $p\ge 2,k,m$. Therefore each such projected intertwiner is zero. For $n=0$, the only numerical boundary channel is $h_{1,-1}=b_1$, so channel exclusion alone does not apply. Creation gives $Y(X,z)\mathbf{1}=e^{zT}X$; $F,H,E$ lie in $S_0$ by its definition and the table $QF=H,QH=E$, so the projection to every quotient factor is zero on the vacuum. For each $r\ge 1$, the primary commutator expresses the projected field on $L(-r)u$ through $L_C(-r)$ applied to its value on $u$, that value, and its derivative; all vanish when the value on $u=\mathbf{1}$ vanishes. Induction on Virasoro descendants therefore gives zero on the entire vacuum factor. Since the quotient is an algebraic direct sum, vanishing of every factor projection means the full image lies in $S_0$. Thus every mode of $F,H,E$ preserves $S_0$; all Virasoro modes preserve it by construction.

To pass from generators to all of $W$, let $D$ be the set of states $a\in V_L$ whose every mode preserves $S_0$. It contains $\mathbf{1},\omega ,F,H,E$. If $a,b\in D$, the module iterate identity expresses each mode of $a_{(r)}b$ as a sum of compositions of modes of $a$ and $b$; on each fixed vector that sum is finite by truncation. Every composition preserves $S_0$, so $a_{(r)}b\in D$ for every $r$. Hence $D$ is a vertex subalgebra containing the four generators, and $W\subset D$. As $\mathbf{1}\in S_0$, every word of generator modes applied to $\mathbf{1}$ lies in $S_0$, giving $W\subseteq S_0$.

Let $U$ be the span of finite words of strictly negative modes of $\omega ,F,H,E$ on $\mathbf{1}$. Write $f_n=e^{-n\alpha }$. The actual exponential product formula in [A1] gives $Y(F,z)f_n=z^{2pn}\exp \text{\hspace{0.17em}⁣}(-{\sum }_{r\ge 1}\frac{x_r}{r}{z}^r)f_{n+1}\mathrm{.}$ Its coefficient of $z^{2pn}$ is $f_{n+1}$, so $f_{n+1}=F_{(-2pn-1)}{f}_n$. Starting with $f_0=\mathbf{1}$, this proves $f_n\in U$ for all $n$.

The derivation identity in [A3] gives $[Q,a_{(m)}]=(Qa{)}_{(m)}$. Commuting $Q$ through a finite word in $U$ and using its table proves $QU\subset U$, hence $Q^j{f}_n\in U$ for every $j,n$. For $r\ge 2$, $L(-r)={\omega }_{(-r+1)}$ is a strictly negative generator mode; for $r=1$, $L(-1)=T$ preserves $U$ because $[T,a_{(-k)}]=k{a}_{(-k-1)}$ for $k\ge 1$ and $T\mathbf{1}=0$. Thus every stipulated Virasoro descendant lies in $U$. The definition of $S_0$ gives $S_0\subseteq U\subseteq W$. With step 2.1 this proves $W=S_0=U$.

Now $S_0=W$ is a vertex subalgebra, contains $\omega ,F,H,E$, and is spanned by the Virasoro descendants of $Q^j{f}_n$ with $f_0=\mathbf{1}$, $f_1=F$, and $f_{n+1}=F_{(-2pn-1)}{f}_n$. These are exactly the hypotheses of [F3], which confirms that the negative-mode span $U$ is the full subalgebra. Hence $W$ is strongly generated by $\omega ,F,H,E$. No equality with a full short-screening kernel was used.