6.5 First-order axioms for waterworld

We summarize the details of how we choose to model WaterWorld boards in first-order logic: exactly what relations we make up, and the formaldomain axioms which capture the game's rules.

This will follow almost exactly the same pattern as our WaterWorld model in propositional logic . However, we will take advantage of the additional flexibility providedby first-order logic.

Rather than modeling only the default 64 WaterWorld board;, we will be able to model any board representable by our relations.This will allow boards of any size and configuration, with one major constrainteach location can have at most three neighboring pirates.

Domain and relations

Our domain is simply the set of all board locations. This set can be arbitrarily largeeven infinite!

The board configuration is given by the binaryneighborrelation $\mathrm{nhbr}$ .

The next relations correspond directly to the propositions in the propositional logic model.

• Whether or not a location contains a pirate: $\mathrm{safe}$ . This is a unary relation.
  We choose not to include a redundant relation $\mathrm{unsafe}$ .
• Unary relations indicating the number of neighboring pirates: $\mathrm{has0}$ , $\mathrm{has1}$ , $\mathrm{has2}$ , and $\mathrm{has3}$ .
  Thus, we have our restriction to three unsafe neighbors. This will also be reflected in our domain axioms below. See also this problem for a discussion of how to avoid this restriction.

In addition, to have encode the domain axioms for an arbitrary domain, we also need an equality relation over our domain of locations.As is traditional, we will use infix notation for this relation, for example, $x=y$ . Furthermore, we will allow ourselves to write $x\neq y$ as shorthand for $\neg (x=y)$ . Thus, we do not need a distinct inequality relation.

Note that these relations describe the state of the underlying boardthe modeland not our particular view of it. Our particular view will be reflected in which formulaswe'll accept as premises. So we'll accept $\mathrm{has2}(A)$ as a premise only when $A$ has been exposed and shows a 2.

The domain axioms

Many of our axioms correspond directly, albeit much more succinctly, with those of the propositional model. In addition, we have axioms that specify that our neighbor and equalityrelations are self-consistent.

Axioms asserting that the neighbor relation is anti-reflexive and symmetric:

• $\forall x\colon \neg \mathrm{nhbr}(x, x)$
• $\forall x\colon \forall y\colon \mathrm{nhbr}(x, y)\implies \mathrm{nhbr}(y, x)$

Axioms asserting that=truly is an equality relation, i.e. , it is reflexive, symmetric, and transitive.

• $\forall x\colon x=x$
• $\forall x\colon \forall y\colon (x=y)\implies (y=x)$
• $\forall x\colon \forall y\colon \forall z\colon ((x=y)\land (y=z))\implies (x=z)$

Axioms asserting that the neighbor counts are correct. Each of these is of the formif location $x$ has $n$ neighboring pirates, then there are $n$ distinct unsafe neighbors of $x$ , and any other distinct neighbor $x$ is safe.We use the equality relation to specify the distinctness of each neighbor.

• $\forall x\colon \mathrm{has0}(x)\implies \forall y\colon \mathrm{nhbr}(x, y)\implies \mathrm{safe}(y)$
• $\forall x\colon \mathrm{has1}(x)\implies \exists a\colon \mathrm{nhbr}(x, a)\land \neg \mathrm{safe}(a)\land \forall y\colon (\mathrm{nhbr}(x, y)\land (a\neq y))\implies \mathrm{safe}(y)$
• $\forall x\colon \mathrm{has2}(x)\implies \exists a\colon \exists b\colon \mathrm{nhbr}(x, a)\land \mathrm{nhbr}(x, b)\land (a\neq b)\land \neg \mathrm{safe}(a)\land \neg \mathrm{safe}(b)\land \forall y\colon (\mathrm{nhbr}(x, y)\land (a\neq y)\land (b\neq y))\implies \mathrm{safe}(y)$
• $\forall x\colon \mathrm{has3}(x)\implies \exists a\colon \exists b\colon \exists c\colon \mathrm{nhbr}(x, a)\land \mathrm{nhbr}(x, b)\land \mathrm{nhbr}(x, c)\land (a\neq b)\land (a\neq c)\land (b\neq c)\land \neg \mathrm{safe}(a)\land \neg \mathrm{safe}(b)\land \neg \mathrm{safe}(c)\land \forall y\colon (\mathrm{nhbr}(x, y)\land (a\neq y)\land (b\neq y)\land (c\neq y))\implies \mathrm{safe}(y)$

In addition, we want the implications to go the opposite way. Otherwise, each of $\mathrm{has0}$ , $\mathrm{has1}$ , $\mathrm{has2}$ , and $\mathrm{has3}$ could always be false, while still satisfying the above!For brevity, we elide the details in the following list:

• $\forall x\colon \forall y\colon \mathrm{nhbr}(x, y)\implies \mathrm{safe}(y)\implies \mathrm{has0}(x)$
• $\forall x\colon \implies \mathrm{has1}(x)$
• $\forall x\colon \implies \mathrm{has2}(x)$
• $\forall x\colon \implies \mathrm{has3}(x)$

Axioms asserting that the neighbor counts are consistent. While redundant, including axioms like the following can be convenient.

• $\forall x\colon \mathrm{has0}(x)\implies \neg (\mathrm{has1}(x)\lor \mathrm{has2}(x)\lor \mathrm{has3}(x))$
• $\forall x\colon \mathrm{has1}(x)\implies \neg (\mathrm{has0}(x)\lor \mathrm{has2}(x)\lor \mathrm{has3}(x))$
• $\forall x\colon \mathrm{has2}(x)\implies \neg (\mathrm{has0}(x)\lor \mathrm{has1}(x)\lor \mathrm{has3}(x))$
• $\forall x\colon \mathrm{has3}(x)\implies \neg (\mathrm{has0}(x)\lor \mathrm{has1}(x)\lor \mathrm{has2}(x))$

Note that this set of axioms is not quite complete, as explored in an exercise .