SlowGuess/NeurIPS_2023_OCR huggingface dataset:
In particular, let  $m$  be a fixed but arbitrary Turing Machine, and  $w$  be a fixed but arbitrary input to be given to machine  $m$ . Then, HALT defines the decision problem that outputs True iff  $m$  halts on input  $w$ .
We construct an oracle for AGENTSGENERATE that can decide HALT as follows. Let  $(\mathcal{A},\mathcal{O})$  be an interface where the observation space is comprised of all configurations of machine  $m$ . Then, we consider a deterministic environment  $e$  that simply produces the next configuration of  $m$  when run on input  $w$ , based on the current tape contents, the state of  $m$ , and the location of the tape head. Note that all three of these elements are contained in a Turing Machine's configuration, and that a single configuration indicates whether the Turing Machine is in a halting state or not. Now, let the action space  $\mathcal{A}$  consist of two actions,  $\{a_{\mathrm{no-op}},a_{\mathrm{halt}}\}$ . On execution of  $a_{\mathrm{no-op}}$  no-op, the environment moves to the next configuration. On execution of  $a_{\mathrm{halt}}$ , the machine halts. That is, we restrict ourselves to the singleton agent set,  $\Lambda$ , containing
the agent  $\lambda^{\circ}$  that outputs  $a_{\mathrm{halt}}$  directly following the machine entering a halting configuration, and  $a_{\mathrm{no-op}}$  otherwise:
$$
\lambda^ {\circ}: h a o \mapsto \left\{ \begin{array}{l l} a _ {\text {h a l t}} & o \text {i s a h a l t i n g c o n f i g r a t i o n}, \\ a _ {\text {n o - o p}} & \text {o t h e r w i s e}. \end{array} \right., \qquad \Lambda = \{\lambda^ {\circ} \}.
$$
Using these ingredients, we take any instance of HALT,  $(m,w)$ , and consider the singleton agent basis:  $\Lambda_{\mathrm{B}}^{\mathrm{I}} = \{a_{\mathrm{no - op}}\}$ .
We make one query to our AGENTSGENERATE oracle, and ask:  $\Lambda_{\mathrm{B}}^{1} \notin \Lambda$ . If it is True, then the histories realizable by  $(\lambda^{\circ}, e)$  pair ensure that the single agent in  $\Lambda$  never emits the  $a_{\mathrm{halt}}$  action, and thus,  $m$  does not halt on  $w$ . If it is False, then there are realizable histories in  $e$  in which  $m$  halts on  $w$ . We thus use the oracle's response directly to decide the given instance of HALT.
B.2.3 Theorem 4.3: Properties of Reaches
We find many similar properties hold for reaches.
1.  $\Leftrightarrow$  and  $\nrightarrow$  are not transitive.
2. Sometimes reaches is not commutative: there exists a pair  $(\Lambda^1, \Lambda^2)$  and  $e \in \mathcal{E}$  such that  $\forall_{\lambda^1 \in \Lambda^1} \lambda^1 \rightsquigarrow \Lambda^2$ , but  $\exists_{\lambda^2 \in \Lambda^2} \lambda^2 \not\rightarrow \Lambda^1$ .
3. For all pairs  $(\Lambda, e)$ , if  $\lambda \in \Lambda$ , then  $\lambda \rightsquigarrow \Lambda$ .
4. Every agent satisfies  $\lambda \stackrel{\wedge}{\rightarrow} \Lambda$  in every environment.
5. The decision problem, Given  $(e,\lambda ,\Lambda)$ , output True iff  $\lambda \rightsquigarrow \Lambda$ , is undecidable.
Again, we prove this result through five lemmas that correspond to each of the above properties.
Many of these properties resemble those in Theorem 4.2. For instance, point (5.) shows that deciding whether a given agent sometimes reaches a basis in an environment is undecidable. We anticipate that the majority of decision problems related to determining properties of arbitrary agent sets will be undecidable, though it is still worth making these arguments carefully. Moreover, there may be interesting special cases in which these decision problems are decidable (and perhaps, efficiently so). Identifying these special cases and their corresponding efficient algorithms is another interesting direction for future work.
Lemma B.9.  $\rightsquigarrow$  and  $\nrightarrow$  are not transitive.
Proof of Lemma B.9.
We construct two counterexamples, one for each of "sometimes reaches"  $(\rightsquigarrow)$  and "never reaches"  $(\not\rightsquigarrow)$ .
Counterexample: Sometimes Reaches. To do so, we begin with a tuple  $(e,\Lambda^1,\Lambda^2,\Lambda^3)$  such that both
$$
\forall_ {\Lambda^ {1} \in \Lambda^ {1}} \lambda^ {1} \stackrel {\varepsilon} {\rightsquigarrow} \Lambda^ {1}, \quad \forall_ {\Lambda^ {2} \in \Lambda^ {2}} \lambda^ {2} \stackrel {\varepsilon} {\rightsquigarrow} \Lambda^ {2}.
$$
We will show that there is an agent,  $\overline{\lambda}^1 \in \Lambda^1$ , such that  $\overline{\lambda}^1 \rightsquigarrow \Lambda^3$ , thus illustrating that sometimes reaches is not guaranteed to be transitive. The basic idea is that sometimes reaches only requires an agent stop its search on one realizable history. So,  $\lambda^1 \rightsquigarrow \Lambda^2$  might happen on some history  $h$ , but each  $\lambda^2 \in \Lambda^2$  might only reach  $\Lambda^3$  on an entirely different history. As a result, reaching  $\Lambda^2$  is not enough to ensure the agent also reaches  $\Lambda^3$ .
In more detail, the agent sets of the counterexample are as follows. Let  $\mathcal{A} = \{a_1, a_2\}$  and  $\mathcal{O} = \{o_1, o_2\}$ . Let  $\Lambda^2$  be all agents that, after ten timesteps, always take  $a_2$ .  $\overline{\lambda}^1$  is simple: it
always takes  $a_1$ , except on one realizable history,  $h^\circ$ , (and all of the realizable successors of  $h^\circ$ ,  $\mathcal{H}_{h^\circ}^{\overline{\lambda},e}$ ), where it switches to taking  $a_2$  after ten timesteps. Clearly  $\overline{\lambda^1} \rightsquigarrow \Lambda^2$ , since after ten timesteps, we know there will be some  $\lambda^2$  such that  $\overline{\lambda}^1(h^\circ h') = \lambda^2(h^\circ h')$  for all realizable history suffixes  $h'$ . Now, by assumption, we know that  $\lambda^2 \rightsquigarrow \Lambda^3$ . This ensures there is a single realizable history  $h$  such that there is an  $\lambda^3$  where  $\lambda^2(hh') = \lambda^3(hh')$  for any realizable suffix  $h'$ . To finish the counterexample, we simply note that this realizable  $h$  can be different from  $h^\circ$  and all of its successors. For example,  $h^\circ$  might be the history containing only  $o_1$  for the first ten timesteps, while  $h$  could be the history containing only  $o_2$  for the first ten timesteps. Thus, this  $\lambda^1$  never reaches  $\Lambda^3$ , and we conclude the counterexample.
Counterexample: Never Reaches. The instance for never reaches is simple: Let  $\mathcal{A} = \{a_1, a_2, a_3\}$ , and  $\Lambda^1 = \Lambda^3$ . Suppose all agents in  $\Lambda^1$  (and thus  $\Lambda^3$ ) only choose actions  $a_1$  and  $a_3$ . Let  $\Lambda^2$  be a singleton,  $\Lambda^2 = \{\lambda^2\}$  such that  $\lambda^2 : h \mapsto a_2$ . Clearly, every  $\lambda^1 \in \Lambda^1$  will never reach  $\Lambda^2$ , since none of them ever choose  $a_2$ . Similarly,  $\lambda^2$  will never reach  $\Lambda^3$ , since no agents in  $\Lambda^3$  choose  $a_2$ . However, by Proposition C.15 and the assumption that  $\Lambda^1 = \Lambda^3$ , we know  $\forall_{\lambda^1 \in \Lambda^1} \lambda^1 \rightsquigarrow \Lambda^3$ . This directly violates transitivity.
This completes the argument for all three cases, and we conclude.
Lemma B.10. Sometimes reaches is not commutative: there exists a pair  $(\Lambda^1, \Lambda^2)$  and  $e \in \mathcal{E}$  such that  $\forall_{\lambda^1 \in \Lambda^1} \lambda^1 \rightsquigarrow \Lambda^2$ , but  $\exists_{\lambda^2 \in \Lambda^2} \lambda^2 \not\Rightarrow \Lambda^1$ .
Proof of Lemma B.10.
The result holds as a straightforward consequence of the following counterexample. Consider the pair of agent sets
$$
\Lambda^ {1} = \{\lambda_ {i}: h \mapsto a _ {1} \}, \qquad \Lambda^ {2} = \{\lambda_ {i}: h \mapsto a _ {1}, \lambda_ {j}: h \mapsto a _ {2} \}.
$$
Note that since  $\lambda_{i}$  is in both sets, and  $\Lambda^1$  is a singleton, we know that  $\lambda \rightsquigarrow \Lambda^1$  in any environment by Lemma B.11. But, clearly  $\lambda_{j}$  never reaches  $\Lambda^1$ , since no agent in  $\Lambda^1$  ever chooses  $a_1$ .
Lemma B.11. For all pairs  $(\Lambda, e)$ , if  $\lambda \in \Lambda$ , then  $\lambda \stackrel{\sim}{\rtimes} \Lambda$ .
Proof of Lemma B.11.
The proposition is straightforward, as any  $\lambda \in \Lambda$  will be equivalent to itself in behavior for all histories.
Lemma B.12. Every agent satisfies  $\lambda \rightsquigarrow \triangleleft$  in every environment.
Proof of Lemma B.12.
This is again a direct consequence of Proposition C.18.
Lemma B.13. The decision problem, AGENTREACHES, Given  $(e,\lambda ,\Lambda)$ , output True iff  $\lambda \rightsquigarrow \Lambda$ , is undecidable.
Proof of Lemma B.13.
We again proceed by reducing AGENTREACHES from the Halting Problem.
We construct an oracle for AGENTREACHES that can decide HALT as follows. Consider the same observation space used in the proof of Lemma B.8: Let  $\mathcal{O}$  be comprised of all configurations of machine  $m$ . Then, sequences of observations are simply evolution of different Turing Machines processing possible inputs. We consider an action space,  $\mathcal{A} = \{a_{\mathrm{halted}}, a_{\mathrm{not - yet}}\}$ , where agents simply report whether the history so far contains a halting configuration.
Then, we consider a deterministic environment  $e$  that simply produces the next configuration of  $m$  when run on input  $w$ , based on the current tape contents, the state of  $m$ , and the location of the tape head. Note again that all three of these elements are contained in a Turing Machine's configuration.
Using these ingredients, we take any instance of HALT,  $(m, w)$ , and build the singleton agent set  $\Lambda_{\mathrm{B}}$  containing only the agent  $\lambda_{\mathrm{halted}}: h \mapsto a_{\mathrm{halted}}$  that always reports the machine as having halted. We then consider whether the agent that outputs  $a_{\mathrm{not-yet}}$  indefinitely until  $m$  reports halting, at which point the agent switches to  $a_{\mathrm{halted}}$ .
We make one query to our AGENTREACHES oracle, and ask:  $\lambda \stackrel{\sim}{\rightsquigarrow} \Lambda_{\mathrm{B}}$ . If it is True, then the branching agent eventually becomes equivalent to  $\lambda_{\mathrm{halted}}$  in that they both indefinitely output  $a_{\mathrm{halted}}$  on at least one realizable history. Since  $e$  is deterministic, we know this equivalence holds across all histories. If the query reports False, then there is no future in  $e$  in which  $m$  halts on  $w$ , otherwise the agent would become equivalent to  $\lambda_{\mathrm{halted}}$ . We thus use the oracle's response directly to decide the given instance of HALT.
C Additional Analysis
Finally, we present a variety of additional results about agents and the generates and reaches operators.
C.1 Additional Analysis: Generates
We first highlight simple properties of the generates operator. Many of our results build around the notion of uniform generation, a variant of the generates operator in which a basis generates an agent set in every environment. We define this operator precisely as follows.
Definition C.1. Let  $\Sigma$  be a set of learning rules over some basis  $\Lambda_{\mathrm{B}}$ . We say that a set  $\Lambda$  is uniformly  $\Sigma$ -generated by  $\Lambda_{\mathrm{B}}$ , denoted  $\Lambda_{\mathrm{B}} \models_{\Sigma} \Lambda$ , if and only if
$$
\forall_ {\lambda \in \Lambda} \exists_ {\sigma \in \Sigma} \forall_ {h \in \mathcal {H}} \lambda (h) = \sigma (h) (h). \tag {C.1}
$$
Definition C.2. We say a basis  $\Lambda_{\mathrm{B}}$  uniformly generates  $\Lambda$ , denoted  $\Lambda_{\mathrm{B}} \models \Lambda$ , if and only if
$$
\exists_ {\Sigma \subseteq \Sigma} \Lambda_ {B} \models_ {\Sigma} \Lambda . \tag {C.2}
$$
We will first show that uniform generation entails generation in a particular environment. As a consequence, when we prove that certain properties hold of uniform generation, we can typically also conclude that the properties hold for generation as well, though there is some subtlety as to when exactly this implication will allow results about  $\models$  to apply directly to  $\varepsilon$ .
Proposition C.1. For any  $(\Lambda_{\mathrm{B}},\Lambda)$  pair, if  $\Lambda_{\mathrm{B}}\models \Lambda$  , then for all  $e\in \mathcal{E}$ $\Lambda_{\mathrm{B}}\vDash \Lambda$
Proof of Proposition C.1.
Recall that in the definition of uniform generation,  $\Lambda_{\mathrm{B}}\models \Lambda$  , we require,
$$
\exists_ {\Sigma \subseteq \Xi} \forall_ {\lambda \in \Lambda} \exists_ {\sigma \in \Sigma} \forall_ {h \in \mathcal {H}} \lambda (h) = \sigma (h) (h). \tag {C.3}
$$
Now, contrast this with generates with respect to a specific environment  $e$ ,
$$
\exists_ {\Sigma \subseteq \mathbb {Z}} \forall_ {\lambda \in \Lambda} \exists_ {\sigma \in \Sigma} \forall_ {h \in \bar {\mathcal {H}}} \lambda (h) = \sigma (h) (h). \tag {C.4}
$$
The only difference in the definitions is that the set of histories quantified over is  $\mathcal{H}$  in the former, and  $\bar{\mathcal{H}} = \mathcal{H}^{\lambda ,e}$  in the latter.
Since  $\tilde{\mathcal{H}}\subseteq \mathcal{H}$  for any choice of environment  $e$  , we can conclude that when Equation C.3, it is also the case that Equation C.4 holds, too. Therefore,  $\Lambda_{\mathrm{B}}\models \Lambda \Rightarrow \Lambda_{\mathrm{B}}\nVdash \Lambda$  for any  $e$
We next show that the subset relation implies generation.
Proposition C.2. Any pair of agent sets  $(\Lambda_{\mathrm{small}},\Lambda_{\mathrm{big}})$  such that  $\Lambda_{\mathrm{small}}\subseteq \Lambda_{\mathrm{big}}$  satisfies
$$
\Lambda_ {\text {b i g}} \vDash \Lambda_ {\text {s m a l l}}. \tag {C.5}
$$
Proof of Proposition C.2.
The result follows from the combination of two facts. First, that all agent sets generate themselves. That is, for arbitrary  $\Lambda$ , we know that  $\Lambda \nsubseteq \Lambda$ , since the trivial set of learning rules,
$$
\Sigma_ {\mathrm {t r}} = \left\{\sigma_ {i}: h \mapsto \lambda_ {i}, \forall_ {\lambda_ {i} \in \Lambda} \right\}, \tag {C.6}
$$
that never switches between agents is sufficient to generate the agent set.
Second, observe that removing an agent from the generated set has no effect on the generates operator. That is, let  $\Lambda' = \Lambda \setminus \lambda$ , for fixed but arbitrary  $\lambda \in \Lambda$ . We see that  $\Lambda \nVdash \Lambda'$ , since  $\Sigma_{\mathrm{tr}}$  is sufficient to generate  $\Lambda'$ , too. By inducting over all removals of agents from  $\Lambda$ , we reach our conclusion.
Next, we establish properties about the sets of learning rules that correspond to the generates operator.
Proposition C.3. For any  $(\Lambda_{\mathrm{B}},\Sigma ,\Lambda)$  such that  $\Lambda_{\mathrm{B}}\models_{\Sigma}\Lambda$  , it holds that
$$
| \Lambda | \leq | \Sigma |. \tag {C.7}
$$
Proof of Proposition C.3.
We proceed toward contradiction, and assume  $|\Lambda| > |\Sigma|$ . Then, there is at least one learning rule  $\sigma \in \Sigma$  that corresponds to two or more distinct agents in  $\Lambda$ . Call this element  $\sigma^{\circ}$ , and without loss of generality let  $\lambda^1$  and  $\lambda^2$  be two distinct agents that are each generated by  $\sigma^{\circ}$  in the sense that
$$
\lambda^ {1} (h) = \sigma^ {\circ} (h) (h), \quad \lambda^ {2} (h) = \sigma^ {\circ} (h) (h), \tag {C.8}
$$
for every  $h \in \mathcal{H}$ . But, by the distinctness of  $\lambda^1$  and  $\lambda^2$ , there must exist a history  $h$  in which  $\lambda^1(h) \neq \lambda^2(h)$ . We now arrive at a contradiction as such a history cannot exist: By Equation C.8, we know that  $\lambda^1(h) = \sigma^\circ(h)(h) = \lambda^2(h)$  for all  $h$ .
We see that the universal learning rules,  $\Sigma$ , is the strongest in the following sense.
Proposition C.4. For any basis  $\Lambda_{\mathrm{B}}$  and agent set  $\Lambda$ , exactly one of the two following properties hold:
1. The agent basis  $\Lambda_{\mathrm{B}}$  uniformly generates  $\Lambda$  under the set of all learning rules:  $\Lambda_{\mathrm{B}} \models_{\Sigma} \Lambda$ .
2. There is no set of learning rules for which the basis  $\Sigma$ -uniformly generates the agent set:  $\neg \exists_{\Sigma \subseteq \Sigma} \Lambda_{\mathrm{B}} \models_{\Sigma} \Lambda$ .
Proof of Proposition C.4.
The proof follows from the law of excluded middle. That is, for any set of learning rules  $\Sigma$ , either it generates  $\Lambda$  or it does not. If it does generate  $\Lambda$ , by Lemma B.6 so does  $\Sigma$ . By consequence, if  $\Sigma$  does not generate  $\Lambda$ , neither do any of its subsets.
Furthermore, uniform generation is also transitive.
Theorem C.5. Uniform generates is transitive: For any triple  $(\Lambda^1, \Lambda^2, \Lambda^3)$ , if  $\Lambda^1 \models \Lambda^2$  and  $\Lambda^2 \models \Lambda^3$ , then  $\Lambda^1 \models \Lambda^3$ .
Proof of Theorem C.5.
Assume  $\Lambda^1 \models \Lambda^2$  and  $\Lambda^2 \models \Lambda^3$ . Then, by Proposition C.4 and the definition of the uniform generates operator, we know that
$$
\forall_ {\lambda^ {2} \in \Lambda^ {2}} \exists_ {\sigma^ {1} \in \Sigma^ {1}} \forall_ {h \in \mathcal {H}} \lambda^ {2} (h) = \sigma^ {1} (h) (h), \tag {C.9}
$$
$$
\forall_ {\lambda^ {3} \in \Lambda^ {3}} \exists_ {\sigma^ {2} \in \Sigma^ {2}} \forall_ {h \in \mathcal {H}} \lambda^ {3} (h) = \sigma^ {2} (h) (h), \tag {C.10}
$$
$$
\forall_ {\lambda^ {2} \in \Lambda^ {2}} \forall_ {h \in \mathcal {H}} \exists_ {\lambda^ {1} \in \Lambda^ {1}} \lambda^ {2} (h) = \lambda^ {1} (h), \tag {C.11}
$$
$$
\forall_ {\lambda^ {3} \in \Lambda^ {3}} \forall_ {h \in \mathcal {H}} \exists_ {\lambda^ {2} \in \Lambda^ {2}} \lambda^ {3} (h) = \lambda^ {2} (h). \tag {C.12}
$$
Then, consider a fixed but arbitrary  $\lambda^3\in \Lambda^3$ . We construct a learning rule defined over  $\Lambda^1$  as  $\sigma^1:\mathcal{H}\to \Lambda^1$  that induces an equivalent agent as follows. For each history,  $h\in \mathcal{H}$ , by Equation C.12 we know that there is an  $\lambda^2$  such that  $\lambda^3 (h) = \lambda^2 (h)$ , and by Equation C.11, there is an  $\lambda^1$  such that  $\lambda^2 (h) = \lambda^1 (h)$ . Then, set  $\sigma^1:h\mapsto \lambda^1$  such that  $\lambda^1 (h) = \lambda^2 (h) = \lambda^3 (h)$ . Since  $h$  and  $\lambda^3$  were chosen arbitrarily, we conclude that
$$
\forall_ {\lambda^ {3} \in \Lambda^ {3}} \forall_ {h \in \mathcal {H}} \exists_ {\lambda^ {1} \in \Lambda^ {1}} \lambda^ {3} (h) = \lambda^ {1} (h).
$$
$$
\forall_ {\lambda^ {3} \in \Lambda^ {3}} \exists_ {\sigma^ {1} \in \Xi^ {1}} \forall_ {h \in \mathcal {H}} \lambda^ {3} (h) = \sigma^ {1} (h) (h).
$$
This is exactly the definition of  $\Sigma$ -uniform generation, and by Proposition C.4, we conclude  $\Lambda^1 \models \Lambda^3$ .
Next, we show that a singleton basis only generates itself.
Proposition C.6. Any singleton basis,  $\Lambda_{\mathrm{B}} = \{\lambda \}$ , only uniformly generates itself.
Proof of Proposition C.6.
Note that generates requires switching between base agents. With only a single agent, there cannot be any switching, and thus, the only agent that can be described as switching amongst the elements of the singleton set  $\Lambda_{\mathrm{B}} = \{\lambda\}$  is the set itself.
C.1.1 Rank and Minimal Bases
As discussed in the paper, one natural reaction to the concept of an agent basis is to ask how we can justify different choices of a basis. And, if we cannot, then perhaps the concept of an agent basis is disruptive, rather than illuminating. In the main text, we suggest that in many situations, the choice of basis is made by the constraints imposed by the problem, such as the available memory. However, there are some objective properties of different bases that can help us to evaluate possible choices of a suitable basis. For instance, some bases are minimal in the sense that they cannot be made smaller while still retaining the same expressive power (that is, while generating the same agent sets). Identifying such minimal sets may be useful, as it is likely that there is good reason to consider only the most compressed agent bases.
To make these intuitions concrete, we introduce the rank of an agent set.
Definition C.3. The rank of an agent set,  $\mathrm{rank}(\Lambda)$ , is the size of the smallest agent basis that uniformly generates it:
$$
\operatorname {r a n k} (\Lambda) = \min  _ {\Lambda_ {B} \subset \mathbb {A}} | \Lambda_ {B} | \quad s. t. \quad \Lambda_ {B} \vDash \Lambda . \tag {C.13}
$$
For example, the agent set,
$$
\Lambda = \left\{\lambda^ {0}: h \mapsto a _ {0}, \right. \tag {C.14}
$$
$$
\lambda^ {1}: h \mapsto a _ {1},
$$
$$
\lambda^ {2}: h \mapsto \left\{ \begin{array}{l l} a _ {0} & | h | \bmod 2 = 0, \\ a _ {1} & | h | \bmod 2 = 1, \end{array} \right.
$$
has  $\mathrm{rank}(\Lambda) = 2$ , since the basis,
$$
\Lambda_ {\mathrm {B}} = \left\{\lambda_ {\mathrm {B}} ^ {0}: h \mapsto a _ {0}, \lambda_ {\mathrm {B}} ^ {1}: h \mapsto a _ {1} \right\},
$$
uniformly generates  $\Lambda$ , and there is no size-one basis that uniformly generates  $\Lambda$  by Proposition C.6.
Using the notion of an agent set's rank, we now introduce the concept of a minimal basis. We suggest that minimal bases are particular important, as they contain no redundancy with respect to their expressive power. Concretely, we define a minimal basis in two slightly different ways depending on whether the basis has finite or infinite rank. In the finite case, we say a basis is minimal if there is no basis of lower rank that generates it.
Definition C.4. An agent basis  $\Lambda_{\mathrm{B}}$  with finite rank is said to be minimal just when there is no smaller basis that generates it,
$$
\forall_ {\Lambda_ {B} ^ {\prime} \subset \mathbb {A}} \Lambda_ {B} ^ {\prime} \vDash \Lambda_ {B} \Rightarrow \operatorname {r a n k} \left(\Lambda_ {B} ^ {\prime}\right) \geq \operatorname {r a n k} \left(\Lambda_ {B}\right). \tag {C.15}
$$
In the infinite case, as all infinite rank bases will have the same effective size, we instead consider a notion of minimiality based on whether any elements can be removed from the basis without changing its expressive power.
Definition C.5. An agent basis  $\Lambda_{\mathrm{B}}$  with infinite rank is said to be minimal just when no proper subset of  $\Lambda_{\mathrm{B}}$  uniformly generates  $\Lambda_{\mathrm{B}}$ .
$$
\forall_ {\Lambda_ {B} ^ {\prime} \subseteq \Lambda_ {B}} \Lambda_ {B} ^ {\prime} \vDash \Lambda_ {B} \Rightarrow \Lambda_ {B} ^ {\prime} = \Lambda_ {B}. \tag {C.16}
$$
Notably, this way of looking at minimal bases will also apply to finite rank agent bases as a direct consequence of the definition of a minimal finite rank basis. However, we still provide both definitions, as a finite rank basis may not contain a subset that generates it, but there may exist a lower rank basis that generates it.
Corollary C.7. As a Corollary of Proposition C.2 and Definition C.4, for any minimal agent basis  $\Lambda_{\mathrm{B}}$ , there is no proper subset of  $\Lambda_{\mathrm{B}}$  that generates  $\Lambda_{\mathrm{B}}$ .
Regardless of whether an agent basis has finite or infinite rank, we say the basis is a minimal basis of an agent set  $\Lambda$  just when the basis uniformly generates  $\Lambda$  and the basis is minimal.
Definition C.6. For any  $\Lambda$ , a minimal basis of  $\Lambda$  is any basis  $\Lambda_{\mathrm{B}}$  that is both (1) minimal, and (2)  $\Lambda_{\mathrm{B}} \models \Lambda$ .
A natural question arises as to whether the minimal basis of any agent set  $\Lambda$  is unique. We answer this question in the negative.
Proposition C.8. The minimal basis of a set of agents is not necessarily unique.
Proof of Proposition C.8.
To prove the claim, we construct an instance of an agent set with two distinct minimal bases. Let  $\mathcal{A} = \{a_0, a_1\}$ , and  $\mathcal{O} = \{o_0\}$ . We consider the agent set containing four agents. The first two map every history to  $a_0$  and  $a_1$ , respectively, while the second two alternate between  $a_0$
and  $a_1$  depending on whether the history is of odd or even length:
$$
\Lambda = \left\{\lambda^ {0}: h \mapsto a _ {0}, \right. \tag {C.17}
$$
$$
\lambda^ {3}: h \mapsto \left\{ \begin{array}{l l} a _ {0} & | h | \bmod 2 = 1, \\ a _ {1} & | h | \bmod 2 = 0, \end{array} \right.
$$
$$
\}.
$$
Note that there are two distinct subsets that each universally generate  $\Lambda$ :