of $i$ and $v$.

3)Suppose that $\langle\dot{a}_\xi:\xi<\lambda^+\rangle$ is a ${{{\mathbb P}^\theta_\lambda}}$–name for a $\lambda^+$–sequence of distinct members of ${\dot{{{\mathbb B}}}^\theta_\lambda}$ and let $p\in{{{\mathbb P}^\theta_\lambda}}$. Applying standard cleaning procedures we find a set $A\subseteq\lambda^+$ of the order type $\theta$, an ordinal $\alpha<
\lambda$ and $\tau^*,n^*,u^*$ and $\langle p_\xi,v_\xi:\xi\in A\rangle$ such that $p\leq p_\xi$, ${{\rm ht}}(p_\xi)=\alpha$, $p_\xi{\Vdash}\dot{a}_\xi=
\tau^*(x_i:i\in v_\xi)$ and $$q\stackrel{\rm def}{=}\langle 0,\tau^*,n^*,u^*,\langle p_\xi,v_\xi:\xi\in
A\rangle\rangle\in P^*_{\alpha+1},$$ where $A$ is identified with $\theta$ by the increasing enumeration (so we will think $A=\theta$). For $\xi<\theta$ let $\tau_\xi=\tau^*(x_i: i\in
v_\xi)\in{{\mathbb B}}^{p_\xi}$. Since $\dot{a}_\xi$ were (forced to be) distinct we know that ${{\mathbb B}}^q\models\tau_\xi\neq\tau_\zeta$ for distinct $\xi,\zeta$. Hence $\tau_\xi\notin\langle x_i:i\in u^*\rangle_{{{\mathbb B}}^{p_\xi}}$ (for each $\xi$) and therefore we may find functions $f^0_\xi,f^1_\xi\in F^{p_\xi}$ such that $f^0_\xi{\restriction}u^*=f^1_\xi{\restriction}u^*$, and $f^0_\xi(\tau_\xi)=0$, $f^1_\xi(\tau_\xi)=1$, and if $\xi<\zeta<\theta$, and $H_{\xi,\zeta}$ is the isomorphism from $p_\xi$ to $p_\zeta$, then $f^\ell_\xi=f^\ell_\zeta{\circ}H_{\xi,\zeta}$. Now fix $\xi<\zeta<\theta$ and let $$g\stackrel{\rm def}{=}\bigcup_{\alpha\leq 3\cdot\xi+2} f^0_\alpha\cup
\bigcup_{3\cdot\xi+2<\alpha<\theta} f^1_\alpha.$$ It should be clear that $g$ is a function from $u^q$ to $2$, and moreover $g\in F^q$. Also easily $$g(\sigma_{\rm maj}(\tau_{3\cdot\xi},\tau_{3\cdot\xi+1},\tau_{3\cdot\xi+ 
2}))=0\ \mbox{ and }\ g(\sigma_{\rm maj}(\tau_{3\cdot\zeta},\tau_{3\cdot
\zeta+1},\tau_{3\cdot\zeta+2}))\}=1.$$ Hence we may conclude that $${{\mathbb B}}^q\models\sigma_{\rm maj}(\tau_{3\cdot\xi},\tau_{3\cdot\xi+1},\tau_{3
\cdot\xi+2})<\sigma_{\rm maj}(\tau_{3\cdot\zeta},\tau_{3\cdot\zeta+1},\tau_{3
\cdot\zeta+2})$$ for $\xi<\zeta<\theta$ (remember the definition of $F^q$ and Proposition \[0.D\]). Consequently we get 