All standard $A_{ij}$ generate $P{B}_n$

Base case. For $n=1$ the index set $\{1\le i<j\le 1\}$ is empty, and the subgroup generated by the empty family is the trivial group, which equals $P{B}_1$ by [F2].

Induction hypothesis. Fix $n\ge 2$ and assume that the open-disc group ${\pi }_1(F_{n-1}(D^{\circ }),Q^{(n-1)})$ is generated by the classes $a_{kl}^{(n-1)}=[z_{W_{kl}}]$ of the standard words at $Q^{(n-1)}$, $1\le k<l\le n-1$; equivalently, by [F4], that $P{B}_{n-1}$ is generated by the classes $A_{kl}$.

Choice, the exact sequence, and the kernel. By [F1] the Axiom of Choice [A1] yields DC, so the short exact sequence of [F2] is available at level $n$ and base configuration $Q^{(n)}$, with truncation $q^{\mathrm{\prime }}=(q_1^{(n)},\dots ,q_{n-1}^{(n)})$: the forgetting map $\phi :P{B}_n\to P{B}_{n-1}$ has $\mathrm{im}\kappa =\ker \phi$ and is surjective, and under the isomorphism ${\iota }_{\ast }^F$ it corresponds to the open-disc forgetting map $p:F_n(D^{\circ })\to F_{n-1}(D^{\circ })$, so $\ker p_{\ast }=({\iota }_{\ast }^F{)}^{-1}(\ker \phi )$. By [F3] the kernel $\ker \phi$ is generated by $A_{1n},\dots ,A_{n-1,n}$, and by [F4] $({\iota }_{\ast }^F{)}^{-1}(A_{in})=(a_{in}^{(n)}{)}^{-1}$; since a subgroup generated by elements equals the one generated by their inverses, $\ker p_{\ast }=⟨a_{1n}^{(n)},\dots ,a_{n-1,n}^{(n)}⟩$.

The affine comparison of the two configurations. Write $h_m=\frac{1}{4(m+1)}$ for the configurations of [F5], so that $h_{n-1}=\frac{n+1}{n}{h}_n$, and for $t\in I$ define the similarity $g_t(z):=(1+\frac{t}{n})z+t(h_{n-1},0)$ of ${\mathbb{R}}^2$, so that $g_0=\mathrm{id}$ and $g_1$ is the scaling $z\mapsto \frac{n+1}{n}z$ followed by the translation by $(h_{n-1},0)$. For $1\le j\le n-1$ one computes $g_1(q_j^{(n)})=\frac{n+1}{n}(2j-n-1)h_n+(h_{n-1},0)=((2j-n-1)h_{n-1}+h_{n-1},0)=((2j-n)h_{n-1},0)=q_j^{(n-1)}$, and the same computation gives $g_1(m_r^{(n)})=m_r^{(n-1)}$ for the midpoints, $1\le r\le n-2$. Since the diamond paths of [F5] are ${\rho }^{(m)}=h_m{\rho }^{(1)}$ for the fixed normalised shape ${\rho }^{(1)}$ given by ${\rho }^{(1)}(t)=(2t-1,-2t)$ for $t\le \frac{1}{2}$ and ${\rho }^{(1)}(t)=(2t-1,2t-2)$ for $t\ge \frac{1}{2}$, one gets $\frac{n+1}{n}{\rho }^{(n)}=h_{n-1}{\rho }^{(1)}={\rho }^{(n-1)}$. Consequently $g_1(m_r^{(n)}\pm {\rho }^{(n)}(s))=g_1(m_r^{(n)})\pm \frac{n+1}{n}{\rho }^{(n)}(s)=m_r^{(n-1)}\pm {\rho }^{(n-1)}(s)$, and the same equality holds for the reflected displacement ${\rho }^{-}$; the similarity $g_1$ has real coefficients and therefore commutes with the reflection $R(x,y)=(x,-y)$ that defines ${\sigma }_r^{-}$ in [F5].

The comparison homotopy. Fix $1\le i<j\le n-1$ (there are no such pairs when $n=2$). Let $z=z_{W_{ij}^{(n)}}$ be the coordinate path of the representative of $W_{ij}$ at $Q^{(n)}$ built from the motions of [F5], a loop in $F_n(D^{\circ })$, and put $u:=[\upsilon ]$ for the loop $\upsilon (s):=(z_1(s),\dots ,z_{n-1}(s))$ in $F_{n-1}(D^{\circ })$, so that $u=p_{\ast }(a_{ij}^{(n)})$; let $\zeta$ be the corresponding coordinate path of $W_{ij}$ at $Q^{(n-1)}$ and $v:=[\zeta ]=a_{ij}^{(n-1)}$. Every strand of the word moves only inside the support discs $U_r$ of the half twists with $r\le n-2$ and off the last strand, so $\mathrm{\mid }{z}_k(s)\mathrm{\mid }\le (n-2)h_n+\frac{3}{2}{h}_n=(n-\frac{1}{2})h_n$ for all $k,s$ (the fixed strands are the base points, of norm at most $(n-1)h_n$). Because each factor of the word and each factor of $\zeta$ is built from the same normalised diamond paths and the midpoints correspond under $g_1$ by step 1.4, and because $g_1$ respects stacking and time reparametrisation, the identities $g_1\circ {\sigma }_r^{(n)}={\sigma }_r^{(n-1)}$, $g_1\circ ({\sigma }_r^{-}{)}^{(n)}=({\sigma }_r^{-}{)}^{(n-1)}$ for $r\le n-2$ imply $g_1(\upsilon (s))=\zeta (s)$ for every $s$. Define $H(s,t):=g_t(\upsilon (s))$. By step 1.4 each $g_t$ is injective, so the $n-1$ coordinates of $H(s,t)$ are pairwise distinct, and $\mathrm{\mid }{H}_k(s,t)\mathrm{\mid }\le (1+\frac{t}{n})(n-\frac{1}{2})h_n+t{h}_{n-1}\le \frac{n+1}{n}(n-\frac{1}{2})h_n+h_{n-1}=\frac{n+1\mathrm{/}2}{4n}<1$, so $H$ takes values in $F_{n-1}(D^{\circ })$ and is continuous; moreover $H(-,0)=\upsilon$, $H(-,1)=\zeta$, and the path $\eta (t):=g_t(q^{\mathrm{\prime }})$ satisfies $H(0,t)=g_t(\upsilon (0))=g_t(q^{\mathrm{\prime }})=\eta (t)=H(1,t)$ because $\upsilon (0)=\upsilon (1)=q^{\mathrm{\prime }}$. Thus $H$ is a homotopy of based $1$-cubes from $\upsilon$ to $\zeta$ whose boundary value is the path $\eta$ in $F_{n-1}(D^{\circ })$ from $q^{\mathrm{\prime }}$ to $Q^{(n-1)}$.

Transporting along the affine path. By step 2.1 the homotopy $H$ has basepoint track $\eta$, so the moving-homotopy transport identity of [F6] in degree one gives $u={\beta }_{\eta }(v)$, that is $p_{\ast }(a_{ij}^{(n)})={\beta }_{\eta }(a_{ij}^{(n-1)})$ for every $1\le i<j\le n-1$, where ${\beta }_{\eta }:{\pi }_1(F_{n-1}(D^{\circ }),Q^{(n-1)})\to {\pi }_1(F_{n-1}(D^{\circ }),q^{\mathrm{\prime }})$ is the isomorphism of [F6].

The images of the older generators generate the quotient. Let $K:=⟨a_{ij}^{(n)}:1\le i<j\le n⟩\le {\pi }_1(F_n(D^{\circ }),Q^{(n)})$ be the subgroup generated by all the raw standard words at level $n$. Since $p_{\ast }$ is a homomorphism and the index set splits into the cases $j\le n-1$ and $j=n$, one has $p_{\ast }(K)=⟨p_{\ast }(a_{ij}^{(n)}):1\le i<j\le n-1⟩=⟨{\beta }_{\eta }(a_{ij}^{(n-1)}):1\le i<j\le n-1⟩={\beta }_{\eta }({\pi }_1(F_{n-1}(D^{\circ }),Q^{(n-1)}))$, where the last equality uses that the classes $a_{ij}^{(n-1)}$ generate the group by the induction hypothesis of step 1.2 and that ${\beta }_{\eta }$ is an isomorphism; hence $p_{\ast }(K)={\pi }_1(F_{n-1}(D^{\circ }),q^{\mathrm{\prime }})$.

Lifting generation to level $n$. By step 1.3 the classes $a_{1n}^{(n)},\dots ,a_{n-1,n}^{(n)}$ lie in $K$ and generate $\ker p_{\ast }$, so $\ker p_{\ast }\subseteq K$. Let $g\in {\pi }_1(F_n(D^{\circ }),Q^{(n)})$. By step 4.1 there is $h\in K$ with $p_{\ast }(h)=p_{\ast }(g)$, hence $g{h}^{-1}\in \ker p_{\ast }\subseteq K$ and therefore $g=(g{h}^{-1})h\in K$. So $K={\pi }_1(F_n(D^{\circ }),Q^{(n)})$.

The closed-disc statement and induction conclusion. Applying the isomorphism ${\iota }_{\ast }^F$ of [F2] to the equality of step 5.1 gives $P{B}_n={\iota }_{\ast }^F(K)=⟨{\iota }_{\ast }^F(a_{ij}^{(n)})⟩$, and by [F4] ${\iota }_{\ast }^F(a_{ij}^{(n)})=A_{ij}^{-1}$, so $P{B}_n=⟨A_{ij}^{-1}:1\le i<j\le n⟩=⟨A_{ij}:1\le i<j\le n⟩$, which is the statement at level $n$; step 1.1 is the base case, so by induction $P{B}_n$ is generated by the standard generators for every $n\ge 1$.