A wf that is true on a finite domain

Question

The wf that I am considering is similar to the one that is in Proving a closed form is true on finite domain. I have read the hints in that post, but I still don't know how to show the following claim:

Let $\mathcal{B}$ be $$ (\forall x)(\forall y)(\forall z) [ A_1^2(x,x)\land (A_1^2(x,y)\land A_1^2(y,z) \Rightarrow A_1^2(x,z) ) \land (A_1^2(x,y) \lor A_1^2(y,x))] \Rightarrow (\exists y)(\forall x) A_1^2(y,x). $$ Then $\mathcal{B}$ is true for all finite domain. In the post that I have linked, it is said that we can use induction. So here is my attempt:

Let $M$ be an interpretation with a domain $D = \{a_1, \dotsc, a_n\}$. We will show that the claim is true using induction on $n$.

For the base case, we have $D = \{a_1\}$, and so there is only one possible sequence $(a_1, a_1, \dotsc) \in \Sigma$. In a case when $(A_1^2)^M = \varnothing$, then $a_1 \notin (A_1^2)^M$ and hence $s$ does not satisfy the antecedent nor the consequent of $\mathcal{B}$. Therefore the implication is true. When $(A_1^2)^M = \{a_1\}$, then the sequence satisfy both the antecedent and the consequent because it is only a sequence of $a_1$. In any case, the wf $\mathcal{B}$ is true for $M$.

Now assume that $\mathcal{B}$ is true in $M$ when $D = \{a_1, \dotsc, a_n\}$ for the induction hypothesis. We extend the interpretation $M$ into $M^{\prime}$ with a domain $D^{\prime} = D \cup \{a_{n+1}\}$. Let $s = (s_1, s_2, \dotsc) \in \Sigma$ be arbitrary.

Case 1: $(A_1^2)^{M^{\prime}} = \varnothing$. Then $\langle s_i, s_j \rangle$ is not in the relation $(A_1^2)^{M^{\prime}}$ for any positive integer $i$ and $j$. The antecedent and the consequent is false and hence, the implication is true. So $\mathcal{B}$ is true for $M^{\prime}$.

Case 2:$(A_1^2)^{M^{\prime}} \neq \varnothing$. All of the atomic wfs in $\mathcal{B}$ are true for $M^{\prime}$ when $\langle s_i, s_j \rangle$ are elements of $D$ by the induction hypothesis.

I got stuck in case 2. How do I show that it is true when $s_i$ and $s_j$ are

1. both $a_{n+1}$
2. one of them are from $D$?

Answer 1

If we have that $D=\{ a_1, \ldots a_n \}$ is the domain of the interpretation $\mathcal M$ with $n$ elements satisfying the formula, this means that we may re-arrange them such that:

$(a_i,a_j) \in (A_1^2)^{\mathcal M}$, for every $i,j$ such that $i \le j$.

Thus, when we add a new element $a_{n+1}$, it is enough to put it "to the right":

$(a_i,a_{n+1}) \in (A_1^2)^{\mathcal M'}$, for every $i \le (n+1)$.