Standard epistemic logic focuses on reasoning about propositional knowledge expressed by knowing that [Hintikka1962]. However, in natural language, various other knowledge expressions are also frequently used, such as knowing what, knowing how, knowing why, and so on.
In particular, knowing how receives much attention in both philosophy and AI. Epistemologists debate about whether knowledge-how is also propositional knowledge [Fantl2008], e.g., whether knowing how to swim can be rephrased using knowing that. In AI, it is crucial to let autonomous agents know how to fulfill certain goals in robotics, game playing, decision making, and multi-agent systems. In fact, a large body of AI planning can be viewed as finding algorithms to let the autonomous planner know how to achieve some propositional goals, i.e., to obtain goal-directed knowledge-how [Gochet2013]. Here, both propositional knowledge and knowledge-how matter, especially in the planning problems where initial uncertainty and non-deterministic actions are present. From a logician’s point of view, it is interesting to see how knowing how interact with knowing that, and how they differ in their reasoning patterns. A logic of knowing how also helps us to find a consistency notion regarding knowledge database with knowing how expressions.
Consider the scenario where a doctor needs a plan to treat a patient and cure his pain (), under the uncertainty about some possible allergy (). If there is no allergy () then simply taking some pills can cure the pain, and the surgery is not an option. On the other hand, in presence of the allergy, the pills may cure the pain or have no effect at all, while the surgery can cure the pain for sure. The model from Figure 1 represents this scenario with an additional action of testing whether . The dotted line represents the initial uncertainty about , and the test on can eliminate this uncertainty (there is no dotted line between and ).
According to the model, to cure the pain (guarantee ) at the end, it makes sense to take the surgery if the result of the test of whether is positive and take the pills otherwise. We can say the doctor in this case knows how to cure the pain.
How to formalize the knowledge-how of the agent in such scenarios with uncertainty? Already since the early days of AI, people have been looking at it in the setting of logics of knowledge and action [McCarthy and Hayes1969, McCarthy1979, Moore1985, Lespérance et al.2000, van der Hoek et al.2000]. However, there has been no consensus on how to capture the logic of “knowing how” formally (cf. the recent surveys [Gochet2013] and [Ågotnes et al.2015]). The difficulties are well discussed in [Jamroga and Ågotnes2007] and [Herzig2015] and simply combining the existing modalities for “knowing that” and “ability” in a logical language like ATEL [van der Hoek and Wooldridge2003] does not lead to a genuine notion of “knowing how”, e.g., knowing how to achieve is not equivalent to knowing that there exists a strategy to make sure . It does not work even when we replace the strategy by uniform strategy where the agent has to choose the same action on indistinguishable states [Jamroga and Ågotnes2007]. Let express that is a way to make sure some goal is achieved, and let be the standard knowledge-that modality. There is a crucial distinction between the de dicto reading of knowing how () and the desired de re reading () endorsed also by linguists and philosophers [Stanley and Williamson2001]. The latter implies the former, but not the other way round. For example, consider a variant of Example 1.1 where no test is available: then the doctor has de dicto knowledge-how to cure, but not the de re one. Proposals to capture the de re reading have been discussed in the literature, such as making the knowledge operator more constructive [Jamroga and Ågotnes2007], making the strategy explicitly specified [Herzig et al.2013, Belardinelli2014], or inserting in-between an existential quantifier and the ability modality in see-to-that-it (STIT) logic [Broersen and Herzig2015].
In [Wang2015, Wang2016b], a new approach is proposed by introducing a single new modality of (conditional) goal-directed knowing how, instead of breaking it down into other modalities. This approach is in line with other de re treatments of non-standard epistemic logics of knowing whether, knowing what and so on (cf. [Wang2016a] for a survey). The semantics of is inspired by the idea of conformant planning based on linear plans [Smith and Weld1998, Yu et al.2016]. It is shown that is not a normal modality, e.g, knowing how to get drunk and knowing how to drive does not entail knowing how to drive when drunk. The work is generalized further in [Li and Wang2017, Li2017]. However, in these previous works, there was no explicit knowing that modality in the language and the semantics of is based on linear plans, which does not capture the broader notion allowing branching plans or strategies that are essential in the scenarios like Example 1.1.
In this paper, we extend this line of work largely in the following aspects:
Both the knowing how modality and knowing that modality are in the language.
Instead of linear plans in [Wang2015], the semantics of our operator is based on strategies (branching plans).
The intuitive idea behind our semantics of is that the agent knows how to achieve iff (s)he has an executable uniform strategy such that the agent knows that:
guarantees in the end given the uncertainty;
always terminates after finitely many steps.
Note that for an agent to know how to make sure , it is not enough to find a plan which works de facto, but the agent should know it works in the end. This is a strong requirement inspired by planning under uncertainty, where the collection of final possible outcomes after executing the plan is required to be a subset of the collection of the goal states [Geffner and Bonet2013].
Technically, our contributions are summarized as follows:
A logical language with both and operators with a semantics which fleshes out formally the above intuitions about knowing how.
A complete axiomatization with intuitive axioms.
Decidability of our logic.
The paper is organized as follows: Section 2 lays out the language and semantics of our framework; Section 3 proposes the axiomatization and proves its soundness; We prove the completeness of our proof system and show the decidability of the logic in Section 4 before we conclude with future work.
2 Language and Semantics
Let PROP be a countable set of propositional symbols.
Definition 2.1 (Language).
The language is defined by the following BNF where :
We use as usual abbreviations and write for .
Definition 2.2 (Models).
A model is a quintuple where:
is a non-empty set,
ACT is a set of actions,
is an equivalence relation on ,
is a binary relation on , and
is a valuation.
Note that the labels in ACT do not appear in the language. The graph in Example 1.1 represents a model with omitted self-loops of (dotted lines), and the equivalence classes induced by are . In this paper we do not require any properties between and to lay out the most general framework. We will come back to particular assumptions like perfect recall at the end of the paper. Given a model and a state , if there exists such that , we say that is executable at . Also note that the actions can be non-deterministic. For each , we use to denote the equivalence class , and use to denote the collection of all the equivalence classes on w.r.t. . We use to indicate that there are and such that . If there is such that , we say is executable at .
Definition 2.3 (Strategies).
Given a model, a (uniformly executable) strategy is a partial function such that is executable at all . Particularly, the empty function is also a strategy, the empty strategy.
Note that the executability is as crucial as uniformity, without which the knowledge-how may be trivialized. We use to denote the domain of . Function can be seen as a binary relation such that implies . Therefore, if is a restriction of , i.e. , it follows that , and for all .
Definition 2.4 (Executions).
Given a strategy w.r.t a model , a possible execution of is a possibly infinite sequence of equivalence classes such that for all . Particularly, is a possible execution if . If the execution is a finite sequence , we call the leaf-node, and an inner-node w.r.t. this execution. If it is infinite, then all are inner-nodes. A possible execution of is complete if it is infinite or its leaf-node is not in .
Given and , we use to denote that extends , i.e., and for all . If , we define . We use to denote the set of all leaf-nodes of all the ’s complete executions (can be many due to non-determinism) starting from , and to denote the set of all the inner-nodes of ’s complete executions starting from . since if is a leaf-node of a complete execution then is not defined at .
Definition 2.5 (Semantics).
Given a pointed model , the satisfaction relation is defined as follows: starting from are finite, where .
Note that the two conditions for in the semantics of reflect our two intuitions mentioned in the introduction. The implicit role of in will become more clear when the axioms are presented. Going back to Example 1.1, we can verify that holds on and due to the strategy . Note that and . On the other hand, is not true on : although the agent can guarantee de facto on by taking a strategy such that and , he cannot know it beforehand since nothing works at to make sure . Readers may also verify that holds at and (hint: a strategy is a partial function).
3.1 The proof system
|all axioms of propositional logic|
Note that we have axioms for . says if is known then you know how to achieve by doing nothing (we allow the empty strategy). reflects the first condition in the semantics that the goal is known after the executions. We will come back to this axiom at the end of the paper. Note that the termination condition is not fully expressible in our language but captures part of it by ruling out strategies that have no terminating executions at all. essentially says that the strategies can be composed. Its validity is quite involved, to which we devote the next subsection. Finally, is the positive introspection axiom for , whose validity is due to uniformity of the strategies on indistinguishable states. The corresponding negative introspection can be derived by using , and :
Note that we do not have the K axiom for . Instead, we have the monotonicity rule . In fact, the logic is not normal, as desired, e.g., is not valid: the existence of two different strategies for different goals does not imply the existence of a unified strategy to realize both goals.
3.2 Validity of
is about the “sequential” compositionality of strategies. Suppose on some pointed model there is a strategy to guarantee that we end up with the states where on each of them we have some other strategy to make sure (). Since the strategies are uniform, we only need to consider some for each . Now to validate , we need to design a unified strategy to compose and those into one strategy to still guarantee (). The general idea is actually simple: first ordering those leafnodes (using Axiom of Choice); then by transfinite induction adjust one by one to make sure these strategies can fit together as a unified strategy ; finally, merge the relevant part of with into the desired strategy. We make this idea precise below. First we need an observation:
Given strategies and with , if and , then a sequence is ’s complete execution from if and only if it is ’s complete execution from .
Left to Right: Let be a ’s complete execution from . We will show it is also a ’s complete execution from . Firstly, we show it is a possible execution give from . If it is not, there exists such that is not the leaf-node of this execution and that . Let be the minimal equivalence class in the sequence with such properties. It follows that and . These are contradictory with .
Next we will show that be a ’s complete execution from . It is obvious if the sequence is infinite. If it is finite, let the leaf-node is . It follows that . Since , it follows . Therefore, the execution is complete given .
Right to Left: Let be a ’s complete execution from , we will show it is also a ’s complete execution from . Since , it is also a possible execution given . If the execution is infinite, it is obvious. If it is finite, let the leaf-node is . It follows that . Since , it follows that . Therefore, the execution is also complete give . ∎
Supposing , we will show that . It follows by the semantics that there exists a strategy such that all ’s complete executions from are finite and for all . If , then , and then it is trivial that . Next we focus on the case of .
According to well-ordering theorem (equivalent to Axiom of Choice), we assume where
is an ordinal number and. Let be an element in ; then . Since for each , it follows that for each there exists a strategy such that all ’s complete executions from are finite and for all . Next, in order to show , we need to define a strategy . The definition consists of the following steps.
Step I. By induction on , we will define a set of strategies where . Let and we define:
We have the following results:
For each , if ;
For each , is a partial function;
For each , where if ;
For each , if is a ’s complete execution from then for some and ;
For each , or .
Proof of claim 3.3.1:
It is obvious.
We prove it by induction on . For the case of , it is obvious. For the case of , it follows by the IH that is a partial function for each . Furthermore, it follows by 1. that for all . Thus, we have is a partial function. Since is a partial function, in order to show is a partial function, we only need to show that . Since , it is obvious that .
We prove it by induction on . It is obvious for the case of . For the case of , given and , we need to show that . Supposing , we will show that , namely . Since and , it follows . Moreover, due to , it follows .
Next, we only need to show . Assuming , it follows that for some . There are two cases: or . If , it follows by the IH that . Contradiction. If , it follows by 1. that . Due to , it follows . It is contradictory with . Thus, we have .
We prove it by induction on . For the case of , due to , it follows that there is a ’s possible execution such that and . Let . (If then ). Since is a ’s complete execution from , it follows that is a ’s complete execution from . It follows by () that is finite. Thus, for some . Since , it follows by that .
For the case of , there are two situations: or . If , it follows that for some . By 3, we have . Since is a ’s complete execution, it follows by Proposition 3.2 that is also a ’s complete execution from . It follows by the IH that for some and .
If , there are two cases: there exist and s.t. , or there do not exist such and . (Please note that due to the fact that is ’s complete execution from ).
for some and some : It follows that is a ’s complete execution from . By 3. and Proposition 3.2, is a ’s complete execution from . By IH, for some and . Therefore, .
If there do not exist and s.t. , it follows that is a ’s possible execution from . Since , then there is a ’s possible execution s.t. , and . Let . (If then ). It follows that is ’s possible execution from . By , all ’s complete executions from are finite. Thus, is finite. Therefore, for some .
We continue to show that . Since is a ’s complete execution from and it is also a ’s possible execution from , there are two cases: for some , or is a ’s complete execution from . If for some , then there exists s.t. and . By IH, . If is a ’s complete execution from , it follows that is a ’s complete execution from . Then by , we have .
If , it follows by that . Otherwise, there are two cases: or . If , it follows by that . Thus, .
If and , we will show that if then . Firstly, we have that . Since , it follows that for some . It follows that there exists such that and . It follows by 4. that .
Step II. We define . It follows by 1. and 2. of Claim 3.3.1 that is indeed a partial function. Then we prove the following claim.
If is a ’s complete execution from then for some and
Proof of claim 3.3.2: Since , it follows that for some . It follows by 5. of Claim 3.3.1 that all ’s complete executions from are finite. Thus, there exists such that for some and is ’s complete execution from . It follows by 5. of Claim 3.3.1 that .
Next, we only need to show . If not, then . We then have that there exists such that . It cannot be that . Otherwise, is not ’s complete execution since by 1. of Claim 3.3.1. Thus, we have . Since we also have that , and , this is contradictory with 3. of Claim 3.3.1. Therefore, we have .
Step III. We define as where and is the strategy mentioned at . Since both and are partial functions, is also a partial function. We then prove the following claim.
If is a ’s complete execution from then for some and .
Proof of claim 3.3.3: Since , there are two cases: or .
If , there are two cases: there exists such that , or there does not exists such . (Please note that due to the fact that is ’s complete execution from ).
If there does not exist s.t. , then is a ’s possible execution from . Since , then there is a ’s possible execution s.t. , and . Let . (If then ). It follows that is ’s possible execution from . By , all ’s complete executions from are finite. Thus, is finite. Therefore, for some .
We continue to show that . Since is a ’s complete execution from and it is also a ’s possible execution from , there are two cases: for some , or is a ’s complete execution from . If for some , it follows by Claim 3.3.2 that . If is a ’s complete execution from , it follows that is a ’s complete execution from . It follows that for some . Since is ’s complete execution from , it follows . We then have , namely . It follows by 5. of Claim 3.3.1 that , namely .
Theorem 3.4 (Soundness).
If then .
4 Completeness and Decidability
Let be a subformula-closed set of formulas. It is obvious that is countable since the whole language itself is countable. Given a set of formulas , let: , , , . Below we define the closure of , and use it to build a canonical model w.r.t. . We will show that when is finite then we can build a finite model.
is defined as:
Definition 4.2 (Atom).
Let . The formula set is an atom of if
or for all ;
Note that if is the whole language then an atom is simply a maximal consistent set. By a standard inductive construction, we can obtain the Lindenbaum-like result in our setting (which is useful to show the existence lemma for ):
Let be an atom of , and . If is consistent then there is an atom of such that , where or .
Let . Since is an atom and , it follows that there is a set such that or for all . Let be all the formulas in . We define as below.
Firstly, we will show is consistent for all . Since is consistent, we only need to show that if is consistent then is consistent, i.e. either or is consistent. Assuming both and are not consistent, it follows that and