If $P$ is a notion of forcing in $M$, then $G$ is a $P$-generic filter over $M$ if $G\subseteq P$ is a filter, and for every $D\in M$ which is a dense subset of $P$, $G\cap D\neq\varnothing$.


Equivalently we can replace $D$ being dense by being pre-dense, open dense, or a maximal antichain.


Given such a generic filter, and $\dot x$ which is a $P$-name, we can define the interpretation of $\dot x$ by the filter $G$ by recursion, $$\dot x^G=\{\dot y^G\mid\exists p\in G:\langle p,\dot y\rangle\in\dot x\}.$$


Clearly, we don't *need* genericity in order to interpret names. We can talk about interpretation using arbitrary filters.



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> **Question.** Given a filter $G$, is there some reasonable condition stating that $G$ is generic if and only if it interprets certain names (e.g. names which are forced to be ordinals) "properly"?
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Genericity by names
Your question is fairly open-ended, but here is one way to do it. For any antichain $A\subset P$, consider the name $\dot a\_A=\{\langle\check a,a\rangle\mid a\in A\}$, which is a mixture of the elements of $A$ on the antichain $A$. This is the name that is trying to be an element of $A$, with values determined by $A$ itself. 


**Theorem.** The following are equivalent, for a filter $G\subset P$.


1. $G$ is $V$-generic.
2. $(\dot a\_A)^G\in A$ for each maximal antichain $A\subset P$.
3. $(\dot a\_A)^G$ is nonempty for every maximal antichain.


Proof. The filter $G$ is generic if and only if it meets every maximal antichain, and in this case, 2 will hold, and clearly 2 implies 3. If 3 holds, then there must be some $a\in A\cap G$, and so $G$ will be $V$-generic. QED


Miha suggests a similar idea in the comments, namely, take any nonempty set $x$, and consider the names $\dot x=\{\langle\check y,a\rangle \mid y\in x\text{ and }a\in A\}$, for any maximal antichain $A$. Miha's observation is that although every condition forces $\check x=\dot x$, nevertheless if $G$ is not generic and misses the antichain $A$, then $\dot x^G$ will be empty. This idea arrives at the following, which seems to meet the requirement of your question.


**Theorem.** (Miha Habič, in comments) The following are equivalent:


1. $G$ is $V$-generic.
2. $G$ interprets all names for non-zero ordinals as non-zero ordinals. That is, if $\mathbb{1}\Vdash(\tau$ is a nonzero ordinal), then $\tau^G$ is a nonzero-ordinal.
3. $G$ interprets all names for the number $1$ correctly.


Proof. Clearly 1 implies 2 and 3. Conversely, if 2 or 3 holds, let $A$ be any maximal antichain and let $\tau=\{\langle 0,a\rangle \mid a\in A\}$. This is a name forced to be the ordinal $1$, and it is easy to see that $\tau^G$ is either $0$ or $1$, depending on whether $G$ meets $A$. So $1$ holds. QED


One can use dense sets instead of antichains, which will avoid a dependence on AC.


Lastly, let me add then when you are entertaining the idea of interpreting names via a filter that is not generic, then you should not be using the value as you have defined it, but rather you should be using the [Boolean ultrapower](http://jdh.hamkins.org/boolean-ultrapowers/) instead. In this method, you use the Boolean algebra rather than just a partial order, and define the equivalence on names $\sigma=\_G\tau\iff[\![\sigma=\tau]\!]\in G$, and similarly for $\sigma\in\_G\tau$. Now, the ultrapower map is $x\mapsto [\check x]\_G$, meaning the equivalence class under this equivalence relation. The result is an elementary embedding $$j:V\to\check V\_G\subseteq V^{\mathbb{B}}/G,$$whose target is the ground model of the corresponding quotient $V^{\mathbb{B}}/G$. The following theorem appears in the Boolean ultrapower paper:


**Theorem.** The following are equivalent:


1. $G\subset P$ is $V$-generic.
2. The Boolean ultrapower by $G$ is an isomorphism of $V$ with $\check V/G$.


This theorem is part of the explanation that "non-generic" is the right generalization of "non-principal" when considering the ultrapower on a complete Boolean algebra instead of merely on a powerset. 


