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First, since I found your notation confusing, I hope you don't mind if I rewrite your Question 1 in more standard notation.

Fix a language $L$. Let $I$ be a non-empty set, and let $(\varphi_i)_{i\in I}$ be a family of $L$-formulas, all of which have free variables from a set $V=\{x_0,x_1,x_2\dots\}$. For each $i\in I$, let $V_i\subseteq V$ be the finite set of variables which are free in $\varphi_i$. We also pick a variable $x_{k_i}\in V$ for each $i\in I$, which may or may not be free in $\varphi_i$.

Let $M$ be an $L$-structure. An assignment $a = (a_x)_{x\in V}$ is a family of elements from $M$ indexed by $V$ (so the variable $x$ gets assigned to $a_x\in M$). Given assignments $a$ and $b$, we write $a\equiv_i b$ if $a_x = b_x$ for all $x\in (V_i\setminus \{x_{k_i}\})$.

Fix an assignment $a$ and an ultrafilter $U$ on $I$.

Statement 1: For all assignments $b$, if $\{ i \mid a \equiv_i b \} \in U$, then $\{ i \mid M\models \varphi_i(b) \} \in U. $

Statement 2: $ \{ i \mid \text{for all assignments } b , \text{if }a \equiv_i b, \text{then } M\models \varphi_i(b)\} \in U.$

Are these two statements equivalent?

Next, to make the logical structure of the statements more transparent, I'll rewrite the statements using the quantifier $\forall^*$, which means for "almost all" $i$ in the sense of the ultrafilter.

Statement 1: $\forall b\, ((\forall^* i\, a\equiv_i b)\rightarrow (\forall ^* i\, M\models \varphi_i(b)))$

Statement 2: $\forall^* i \forall b\, (a\equiv_i b\rightarrow M\models \varphi_i(b))$

Now the nice thing about the ultrafilter $\forall^*$ quantifier is that it commutes with / distributes over all propositional connectives. So Statement 1 is equivalent to $\forall b\forall^* i\,(a\equiv_i b\rightarrow M\models \varphi_i(b))$.

Unfortunately, the quantifier $\forall^* i$ does not commute with $\forall b$. Statement 2 obviously implies Statement 1, since $\forall^* i\forall b$ is stronger than $\forall b\forall^* i$. But the converse may not hold.

I'll give a counterexample to the implication from Statement 1 to Statement 2.

Let $L$ be the empty language (the language of equality). Let $I = \omega$ and $U$ any nonprincipal ultrafilter on $\omega$.

For each $i\in \omega$, let $\varphi_i$ be $\bigwedge_{0\leq j<k\leq i+1} x_j = x_k$. Note that the set of free variables in $\varphi_i$ is $V_i = \{x_0,\dots,x_{i+1}\}$. Let $k_i = i+1$, so the specified variable $x_{k_i}$ is $x_{i+1}$.

Let $M$ be any set with at least two elements, and let $0$ and $1$ be distinct elements of $M$. Let $a$ be the constant assignment with value $0$, i.e., $a_k = 0$ for all $k$. Note that for an assignment $b$ and $i\in \omega$, $a\equiv_i b$ if and only if $b_j = a_j = 0$ for all $j\leq i$ (since $V_i\setminus \{x_{k_i}\} = \{x_0,\dots,x_i\}$).

Now with this setup, Statement 1 is true. Let $b$ be an assignment, and suppose $\{i\mid a\equiv_i b\}\in U$. I claim that $b = a$. Indeed, for all $j\in \omega$, since $\{i\mid a\equiv_i b\}$ is infinite, there exists $i\geq j$ such that $a\equiv_i b$, and hence $b_j = a_j$.Thus $b$ is the constant assignment with value $0$, so $\{i\mid M\models \varphi_i(b)\} = \omega \in U$.

But Statement 2 is false. Fix $i\in \omega$, and let $b$ be the assignment with $b_j = 0$ for all $j\leq i$ and $b_j = 1$ for all $j>i$. Then $a\equiv_i b$, but $M\not\models \varphi_i b$, since $b_{i+1} = 1 \neq 0 = b_i$.

Thus $\{ i \mid \text{for all assignments } b , \text{if }a \equiv_i b, \text{then } M\models \varphi_i(b)\} = \varnothing \notin U$.

Regarding Question 2, you probably know that if $U$ and $V$ are ultrafilters on $X$, then $U\otimes V = \{Y\subseteq X^2\mid \{i\mid \{j\mid (i,j)\in Y\}\in V\}\in U\}$ is an ultrafilter on $X^2$. So you can rewrite your statement as $\{(i,j)\mid \phi(i,j)\}\cap \{(j,i)\mid \phi(i,j)\}\in U\otimes U$. But this is just different notation. I doubt there is a conceptually simpler way to describe the concept.

First, since I found your notation confusing, I hope you don't mind if I rewrite your Question 1 in more standard notation.

Fix a language $L$. Let $I$ be a non-empty set, and let $(\varphi_i)_{i\in I}$ be a family of $L$-formulas, all of which have free variables from a set $V=\{x_0,x_1,x_2\dots\}$. For each $i\in I$, let $V_i\subseteq V$ be the finite set of variables which are free in $\varphi_i$. We also pick a variable $x_{k_i}\in V$ for each $i\in I$, which may or may not be free in $\varphi_i$.

Let $M$ be an $L$-structure. An assignment $a = (a_x)_{x\in V}$ is a family of elements from $M$ indexed by $V$ (so the variable $x$ gets assigned to $a_x\in M$). Given assignments $a$ and $b$, we write $a\equiv_i b$ if $a_x = b_x$ for all $x\in (V_i\setminus \{x_{k_i}\})$.

Fix an assignment $a$ and an ultrafilter $U$ on $I$.

Statement 1: For all assignments $b$, if $\{ i \mid a \equiv_i b \} \in U$, then $\{ i \mid M\models \varphi_i(b) \} \in U. $

Statement 2: $ \{ i \mid \text{for all assignments } b , \text{if }a \equiv_i b, \text{then } M\models \varphi_i(b)\} \in U.$

Are these two statements equivalent?

Next, to make the logical structure of the statements more transparent, I'll rewrite the statements using the quantifier $\forall^*$, which means for "almost all" $i$ in the sense of the ultrafilter.

Statement 1: $\forall b\, ((\forall^* i\, a\equiv_i b)\rightarrow (\forall ^* i\, M\models \varphi_i(b)))$

Statement 2: $\forall^* i \forall b\, (a\equiv_i b\rightarrow M\models \varphi_i(b))$

Now the nice thing about the ultrafilter $\forall^*$ quantifier is that it commutes with / distributes over all propositional connectives. So Statement 1 is equivalent to $\forall b\forall^* i\,(a\equiv_i b\rightarrow M\models \varphi_i(b))$.

Unfortunately, the quantifier $\forall^* i$ does not commute with $\forall b$. Statement 2 obviously implies Statement 1, since $\forall^* i\forall b$ is stronger than $\forall b\forall^* i$. But the converse may not hold.

I'll give a counterexample to the implication from Statement 1 to Statement 2.

Let $L$ be the empty language (the language of equality). Let $I = \omega$ and $U$ any nonprincipal ultrafilter on $\omega$.

For each $i\in \omega$, let $\varphi_i$ be $\bigwedge_{0\leq j<k\leq i+1} x_j = x_k$. Note that the set of free variables in $\varphi_i$ is $V_i = \{x_0,\dots,x_{i+1}\}$. Let $k_i = i+1$, so the specified variable $x_{k_i}$ is $x_{i+1}$.

Let $M$ be any set with at least two elements, and let $0$ and $1$ be distinct elements of $M$. Let $a$ be the constant assignment with value $0$, i.e., $a_k = 0$ for all $k$. Note that for an assignment $b$ and $i\in \omega$, $a\equiv_i b$ if and only if $b_j = a_j = 0$ for all $j\leq i$ (since $V_i\setminus \{x_{k_i}\} = \{x_0,\dots,x_i\}$).

Now with this setup, Statement 1 is true. Let $b$ be an assignment, and suppose $\{i\mid a\equiv_i b\}\in U$. I claim that $b = a$. Indeed, for all $j\in \omega$, since $\{i\mid a\equiv_i b\}$ is infinite, there exists $i\geq j$ such that $a\equiv_i b$, and hence $b_j = a_j = 0$. Thus $b$ is the constant assignment with value $0$, so $\{i\mid M\models \varphi_i(b)\} = \omega \in U$.

But Statement 2 is false. Fix $i\in \omega$, and let $b$ be the assignment with $b_j = 0$ for all $j\leq i$ and $b_j = 1$ for all $j>i$. Then $a\equiv_i b$, but $M\not\models \varphi_i(b)$, since $b_{i+1} = 1 \neq 0 = b_i$.

Thus $\{ i \mid \text{for all assignments } b , \text{if }a \equiv_i b, \text{then } M\models \varphi_i(b)\} = \varnothing \notin U$.

Regarding Question 2, you probably know that if $U$ and $V$ are ultrafilters on $X$, then $U\otimes V = \{Y\subseteq X^2\mid \{i\mid \{j\mid (i,j)\in Y\}\in V\}\in U\}$ is an ultrafilter on $X^2$. So you can rewrite your statement as $\{(i,j)\mid \phi(i,j)\}\cap \{(j,i)\mid \phi(i,j)\}\in U\otimes U$. But this is just different notation. I doubt there is a conceptually simpler way to describe the concept.

This is not really an answer to the question, just an attempt to sort it out. First, since I find your notation bizarre, I hope you don't mind if I rewrite your Question 1 in more standard notation.

Is the above an accurate understanding of your question?

Next, to make the logical structure of the statements more transparent, I'll rewrite the statements using the quantifier $\forall^*$, which means for "almost all" $i$ in the sense of the ultrafilter.

Now the nice thing about the $\forall^*$ quantifier is that it commutes with / distributes over all propositional connectives. So Statement 1 is equivalent to $\forall b\forall^* i\,(a\equiv_i b\rightarrow M\models \varphi_i(b))$.

Now it's possible that in your case, the two statements happen to be equivalent based on their actual content - but the statements seem rather unnatural to me, so I don't feel like thinking too hard about it before you confirm that I've actually understood the statements correctly. Maybe you could say something about the motivation for considering these statements?

First, since I found your notation confusing, I hope you don't mind if I rewrite your Question 1 in more standard notation.

Next, to make the logical structure of the statements more transparent, I'll rewrite the statements using the quantifier $\forall^*$, which means for "almost all" $i$ in the sense of the ultrafilter.

Now the nice thing about the ultrafilter $\forall^*$ quantifier is that it commutes with / distributes over all propositional connectives. So Statement 1 is equivalent to $\forall b\forall^* i\,(a\equiv_i b\rightarrow M\models \varphi_i(b))$.

I'll give a counterexample to the implication from Statement 1 to Statement 2.

Let $L$ be the empty language (the language of equality). Let $I = \omega$ and $U$ any nonprincipal ultrafilter on $\omega$.

For each $i\in \omega$, let $\varphi_i$ be $\bigwedge_{0\leq j<k\leq i+1} x_j = x_k$. Note that the set of free variables in $\varphi_i$ is $V_i = \{x_0,\dots,x_{i+1}\}$. Let $k_i = i+1$, so the specified variable $x_{k_i}$ is $x_{i+1}$.

Let $M$ be any set with at least two elements, and let $0$ and $1$ be distinct elements of $M$. Let $a$ be the constant assignment with value $0$, i.e., $a_k = 0$ for all $k$. Note that for an assignment $b$ and $i\in \omega$, $a\equiv_i b$ if and only if $b_j = a_j = 0$ for all $j\leq i$ (since $V_i\setminus \{x_{k_i}\} = \{x_0,\dots,x_i\}$).

Now with this setup, Statement 1 is true. Let $b$ be an assignment, and suppose $\{i\mid a\equiv_i b\}\in U$. I claim that $b = a$. Indeed, for all $j\in \omega$, since $\{i\mid a\equiv_i b\}$ is infinite, there exists $i\geq j$ such that $a\equiv_i b$, and hence $b_j = a_j$.Thus $b$ is the constant assignment with value $0$, so $\{i\mid M\models \varphi_i(b)\} = \omega \in U$.

But Statement 2 is false. Fix $i\in \omega$, and let $b$ be the assignment with $b_j = 0$ for all $j\leq i$ and $b_j = 1$ for all $j>i$. Then $a\equiv_i b$, but $M\not\models \varphi_i b$, since $b_{i+1} = 1 \neq 0 = b_i$.

Thus $\{ i \mid \text{for all assignments } b , \text{if }a \equiv_i b, \text{then } M\models \varphi_i(b)\} = \varnothing \notin U$.

This is not really an answer to the question, just an attempt to sort it out. First, since I find your notation bizarre, I hope you don't mind if I rewrite your Question 1 in more standard notation.

Fix a language $L$. Let $I$ be a non-empty set, and let $(\varphi_i)_{i\in I}$ be a family of $L$-formulas, all of which have free variables from a set $V=\{x_0,x_1,x_2\dots\}$. For each $i\in I$, let $V_i\subseteq V$ be the finite set of variables which are free in $\varphi_i$. We also pick a variable $x_{k_i}\in V$ for each $i\in I$, which may or may not be free in $\varphi_i$.

Let $M$ be an $L$-structure. An assignment $a = (a_x)_{x\in V}$ is a family of elements from $M$ indexed by $V$ (so the variable $x$ gets assigned to $a_x\in M$). Given assignments $a$ and $b$, we write $a\equiv_i b$ if $a_x = b_x$ for all $x\in (V_i\setminus \{x_{k_i}\})$.

Fix an assignment $a$ and an ultrafilter $U$ on $I$.

Statement 1: For all assignments $b$, if $\{ i \mid a \equiv_i b \} \in U$, then $\{ i \mid M\models \varphi_i(b) \} \in U. $

Statement 2: $ \{ i \mid \text{for all assignments } b , \text{if }a \equiv_i b, \text{then } M\models \varphi_i(b)\} \in U.$

Are these two statements equivalent?

Is the above an accurate understanding of your question?

Next, to make the logical structure of the statements more transparent, I'll rewrite the statements using the quantifier $\forall^*$, which means for "almost all" $i$ in the sense of the ultrafilter.

Statement 1: $\forall b\, ((\forall^* i\, a\equiv_i b)\rightarrow (\forall ^* i\, M\models \varphi_i(b)))$

Statement 2: $\forall^* i \forall b\, (a\equiv_i b\rightarrow M\models \varphi_i(b))$

Now the nice thing about the $\forall^*$ quantifier is that it commutes with / distributes over all propositional connectives. So Statement 1 is equivalent to $\forall b\forall^* i\,(a\equiv_i b\rightarrow M\models \varphi_i(b))$.

Unfortunately, the quantifier $\forall^* i$ does not commute with $\forall b$. Statement 2 obviously implies Statement 1, since $\forall^* i\forall b$ is stronger than $\forall b\forall^* i$. But the converse may not hold.

Now it's possible that in your case, the two statements happen to be equivalent based on their actual content - but the statements seem rather unnatural to me, so I don't feel like thinking too hard about it before you confirm that I've actually understood the statements correctly.

Regarding Question 2, you probably know that if $U$ and $V$ are ultrafilters on $X$, then $U\otimes V = \{Y\subseteq X^2\mid \{i\mid \{j\mid (i,j)\in Y\}\in V\}\in U\}$ is an ultrafilter on $X^2$. So you can rewrite your statement as $\{(i,j)\mid \phi(i,j)\}\cap \{(j,i)\mid \phi(i,j)\}\in U\otimes U$. But this is just different notation. I doubt there is a conceptually simpler way to describe the concept.

This is not really an answer to the question, just an attempt to sort it out. First, since I find your notation bizarre, I hope you don't mind if I rewrite your Question 1 in more standard notation.

Fix a language $L$. Let $I$ be a non-empty set, and let $(\varphi_i)_{i\in I}$ be a family of $L$-formulas, all of which have free variables from a set $V=\{x_0,x_1,x_2\dots\}$. For each $i\in I$, let $V_i\subseteq V$ be the finite set of variables which are free in $\varphi_i$. We also pick a variable $x_{k_i}\in V$ for each $i\in I$, which may or may not be free in $\varphi_i$.

Let $M$ be an $L$-structure. An assignment $a = (a_x)_{x\in V}$ is a family of elements from $M$ indexed by $V$ (so the variable $x$ gets assigned to $a_x\in M$). Given assignments $a$ and $b$, we write $a\equiv_i b$ if $a_x = b_x$ for all $x\in (V_i\setminus \{x_{k_i}\})$.

Fix an assignment $a$ and an ultrafilter $U$ on $I$.

Statement 1: For all assignments $b$, if $\{ i \mid a \equiv_i b \} \in U$, then $\{ i \mid M\models \varphi_i(b) \} \in U. $

Statement 2: $ \{ i \mid \text{for all assignments } b , \text{if }a \equiv_i b, \text{then } M\models \varphi_i(b)\} \in U.$

Are these two statements equivalent?

Is the above an accurate understanding of your question?

Next, to make the logical structure of the statements more transparent, I'll rewrite the statements using the quantifier $\forall^*$, which means for "almost all" $i$ in the sense of the ultrafilter.

Statement 1: $\forall b\, ((\forall^* i\, a\equiv_i b)\rightarrow (\forall ^* i\, M\models \varphi_i(b)))$

Statement 2: $\forall^* i \forall b\, (a\equiv_i b\rightarrow M\models \varphi_i(b))$

Now the nice thing about the $\forall^*$ quantifier is that it commutes with / distributes over all propositional connectives. So Statement 1 is equivalent to $\forall b\forall^* i\,(a\equiv_i b\rightarrow M\models \varphi_i(b))$.

Unfortunately, the quantifier $\forall^* i$ does not commute with $\forall b$. Statement 2 obviously implies Statement 1, since $\forall^* i\forall b$ is stronger than $\forall b\forall^* i$. But the converse may not hold.

Now it's possible that in your case, the two statements happen to be equivalent based on their actual content - but the statements seem rather unnatural to me, so I don't feel like thinking too hard about it before you confirm that I've actually understood the statements correctly. Maybe you could say something about the motivation for considering these statements?

Regarding Question 2, you probably know that if $U$ and $V$ are ultrafilters on $X$, then $U\otimes V = \{Y\subseteq X^2\mid \{i\mid \{j\mid (i,j)\in Y\}\in V\}\in U\}$ is an ultrafilter on $X^2$. So you can rewrite your statement as $\{(i,j)\mid \phi(i,j)\}\cap \{(j,i)\mid \phi(i,j)\}\in U\otimes U$. But this is just different notation. I doubt there is a conceptually simpler way to describe the concept.