2 added 28 characters in body

Yes. In fact, to one can prove Moser and Tardos' the refinement one adapts stated in your question by a slight adaptation of the proof of the classical case.

That is

More precisely, the proof is by recursion over the number of events in $\mathbf{A}$. Like in the classical case, assuming that the statement holds for every collection of $n$ events, it is enough to show that $P(A|\bar{B})\le x(A)$ for every collection $\mathbf{A}=${A}$\cup${$A_i;1\le i\le n$}, where $B$ denotes the union of the events $A_i$ for $1\le i\le n$ and $\bar B$ denotes the complement of $B$.

The first step is to decompose $B$ into $B=C\cup F$ where $C$ is the union of the events in $\Gamma(A)$ and $F$ the union of the events not in $\Gamma(A)$. Hence, $P(A|\bar{B})=P(A|\bar{C},\bar{F})=N/D$ where $N=P(A,\bar{C}|\bar{F})$ and $D=P(\bar{C}|\bar{F})$.

As regards the numerator, $N\le P(A|\bar{F})=P(A)$, where the equality stems from the fact that $A$ is independent of the $A_i$ not in $\Gamma(A)$, which define $F$.

Some additional work is required to deal with the denumerator. Assume wlog that $\Gamma(A)=${$A_i;1\le i\le q$} and write $\bar C$ as $\bar C=\bar A_1\cap \bar A_2\cap\cdots\cap \bar A_q$. Then, by Bayes formula, $P(\bar C|\bar{F})$ is the product over $1\le i\le q$ of $P(\bar A_i|\bar C_i,\bar{F})$, where each $C_i$ is the union of $A_j$ for $j\le q$, $j\ne i$. Now, each of these probabilities is conditional on an event which involves at most $q-1\le n$ events in $\mathbf A$ hence $P(\bar A_i|\bar C_i,\bar{F})\ge1-x(A_i)$ by the recursion hypothesis applied to the collection {$A_i;1\le i\le n$}.

Finally, $N\le P(A)\le x(A)\prod_{i\le q}(1-x(A_i))$ and $D\ge\prod_{i\le q}(1-x(A_i))$ hence $P(A|\bar{B})\le x(A)$ for every collection $\mathbf A$ of size $n+1$ as above, which proves the theorem.

1

Yes. In fact, to prove Moser and Tardos' refinement one adapts the proof of the classical case.

That is, the proof is by recursion over the number of events in $\mathbf{A}$. Like in the classical case, assuming that the statement holds for every collection of $n$ events, it is enough to show that $P(A|\bar{B})\le x(A)$ for every collection $\mathbf{A}=${A}$\cup${$A_i;1\le i\le n$}, where $B$ denotes the union of the events $A_i$ for $1\le i\le n$ and $\bar B$ denotes the complement of $B$.

The first step is to decompose $B$ into $B=C\cup F$ where $C$ is the union of the events in $\Gamma(A)$ and $F$ the union of the events not in $\Gamma(A)$. Hence, $P(A|\bar{B})=P(A|\bar{C},\bar{F})=N/D$ where $N=P(A,\bar{C}|\bar{F})$ and $D=P(\bar{C}|\bar{F})$.

As regards the numerator, $N\le P(A|\bar{F})=P(A)$, where the equality stems from the fact that $A$ is independent of the $A_i$ not in $\Gamma(A)$, which define $F$.

Some additional work is required to deal with the denumerator. Assume wlog that $\Gamma(A)=${$A_i;1\le i\le q$} and write $\bar C$ as $\bar C=\bar A_1\cap \bar A_2\cap\cdots\cap \bar A_q$. Then, by Bayes formula, $P(\bar C|\bar{F})$ is the product over $1\le i\le q$ of $P(\bar A_i|\bar C_i,\bar{F})$, where each $C_i$ is the union of $A_j$ for $j\le q$, $j\ne i$. Now, each of these probabilities is conditional on an event which involves at most $q-1\le n$ events in $\mathbf A$ hence $P(\bar A_i|\bar C_i,\bar{F})\ge1-x(A_i)$ by the recursion hypothesis applied to the collection {$A_i;1\le i\le n$}.

Finally, $N\le P(A)\le x(A)\prod_{i\le q}(1-x(A_i))$ and $D\ge\prod_{i\le q}(1-x(A_i))$ hence $P(A|\bar{B})\le x(A)$ for every collection $\mathbf A$ of size $n+1$ as above, which proves the theorem.