The collection of all formulas $\varphi$ as above forms a directed set (with $\varphi_1\leq \varphi_2$ iff $\varphi_2 \vdash \varphi_1$) and similarly the collection of all $\delta$’s from $p$ forms a directed set; the product of these two directed sets is also a directed set with the product preorder, and the limits below are computed with respect to this product preorder.

By (\[eq: 2\]), we conclude that

$$\lim_{\delta,\varphi} g_{\delta,\varphi}  (q/E_\Delta)=\operatorname{{tp}}_{\Delta}(ab/M)/E_\Delta.\label{eq: 3}$$

On the other hand, $$\lim_{\delta,\varphi} \operatorname{{tp}}_{\Delta}(g_{\delta,\varphi}/M) = p,$$ so $$\lim_{\delta,\varphi} \operatorname{{tp}}_{\Delta}(g_{\delta,\varphi}/M)/E_\Delta = p/E_\Delta,$$ which by virtue of (\[eq: 0\]) implies that

$$\lim_{\delta,\varphi} g_{\delta,\varphi} (q/E_\Delta) = p/E_\Delta * q/E_\Delta. \label{eq: 4}$$

From (\[eq: 3\]) and (\[eq: 4\]), we get $p/E_\Delta * q/E_\Delta = \operatorname{{tp}}_{\Delta}(ab/M)/E_\Delta$.

Define a relation ${F_0}_\Delta$ on $G^*$ as follows. ${F_0}_\Delta(\alpha,\beta)$ holds if there exist $a, b, a_1, b_1 \in G^*$ such that the following conditions hold:

1.  $\alpha=a\cdot b$ and $\beta=a_1\cdot b_1$,

2.  $\operatorname{{tp}}_{\Delta}(a/M)=\operatorname{{tp}}_{\Delta}(a_1/M)$ and $\operatorname{{tp}}_{\Delta}(b/M)=\operatorname{{tp}}_{\Delta}(b_1/M)$,

3.  $\operatorname{{tp}}_{\Delta}(a/M,b)$ and $\operatorname{{tp}}_{\Delta}(a_1/M,b_1)$ are both finitely satisfiable in $M$.

The relation ${F_0}_\Delta$ is clearly reflexive and symmetric, but there is no obvious reason why it should be transitive. Let $F_\Delta$ be the transitive closure of ${F_0}_\Delta$. We see that ${F_0}_\Delta$ and $F_\Delta$ are $M$-invariant. Finally, let ${\bar F}_\Delta$ be the finest 