6.4 Propositional axioms for waterworld

We summarize the details of how we choose to model WaterWorld boardsin propositional logic: exactly what propositions we make up,and the formal domain axioms which capture the game's rules.

The board is fixed at 64, named $A$ ,, $Z$ (with I and O omitted).

Propositions

There are a myriad of propositions for WaterWorld, which can be grouped:

• Whether or not a location contains a pirate: $\mathrm{A-unsafe}$ , $\mathrm{B-unsafe}$ ,, $\mathrm{Z-unsafe}$ .
• Whether or not a location contains no pirate: $\mathrm{A-safe}$ , $\mathrm{B-safe}$ ,, $\mathrm{Z-safe}$ .
  Yes, using the intended interpretation, these are redundant with the previous ones.Some domain axioms below will formalize this.
• Propositions indicating the number of neighboring pirates, to a location: $\mathrm{A-has-0}$ , $\mathrm{A-has-1}$ , $\mathrm{A-has-2}$ , $\mathrm{B-has-0}$ , $\mathrm{B-has-1}$ , $\mathrm{B-has-2}$ ,, $\mathrm{H-has-0}$ , $\mathrm{H-has-1}$ , $\mathrm{H-has-2}$ , $\mathrm{H-has-3}$ ,, $\mathrm{Z-has-0}$ , $\mathrm{Z-has-1}$ . These are all true/false propositions;there are no explicit numbers in the logic.A domain axiom below will assert that whenever (say) $\mathrm{B-has-1}$ is true, then $\mathrm{B-has-0}$ and $\mathrm{B-has-2}$ are both false.
  There is no proposition $\mathrm{A-has-3}$ since location $A$ has only two neighbors. Similarly, there is no proposition $\mathrm{B-has-3}$ . We could have chosen to include those, but under the intended interpretationthey'd always be false.

These propositions describe the state of the underlying boardthe modeland not our particular view of it. Our particular view will be reflected in which formulas we'll accept as premises. So we'll accept $\mathrm{A-has-2}$ as a premise only when $A$ has been exposed and shows a 2.

The domain axioms

Axioms asserting that the neighbor counts are correct:

• Count of 0:
  • A0: $\mathrm{A-has-0}\implies (\mathrm{B-safe}\land \mathrm{G-safe})$
  • H0: $\mathrm{H-has-0}\implies (\mathrm{G-safe}\land \mathrm{J-safe}\land \mathrm{P-safe})$
  • Z0: $\mathrm{Z-has-0}\implies (\mathrm{Y-safe})$
• Count of 1:
  • A1: $\mathrm{A-has-1}\implies (\mathrm{B-safe}\land \mathrm{G-unsafe})\lor (\mathrm{B-unsafe}\land \mathrm{G-safe})$
  • H1: $\mathrm{H-has-1}\implies (\mathrm{G-safe}\land \mathrm{J-safe}\land \mathrm{P-unsafe})\lor (\mathrm{G-safe}\land \mathrm{J-unsafe}\land \mathrm{P-safe})\lor (\mathrm{G-unsafe}\land \mathrm{J-safe}\land \mathrm{P-safe})$
  • Z1: $\mathrm{Z-has-1}\implies (\mathrm{Y-unsafe})$
• Count of 2:
  • A2: $\mathrm{A-has-2}\implies (\mathrm{B-unsafe}\land \mathrm{G-unsafe})$
  • H2: $\mathrm{H-has-2}\implies (\mathrm{G-safe}\land \mathrm{J-unsafe}\land \mathrm{P-unsafe})\lor (\mathrm{G-unsafe}\land \mathrm{J-safe}\land \mathrm{P-unsafe})\lor (\mathrm{G-unsafe}\land \mathrm{J-unsafe}\land \mathrm{P-safe})$
  There aren't any such axioms for locations with only one neighbor.
• Count of 3:
  • H3: $\mathrm{H-has-3}\implies (\mathrm{G-unsafe}\land \mathrm{J-unsafe}\land \mathrm{P-unsafe})$
  There aren't any such axioms for locations with only one or two neighbors.

Axioms asserting that the propositions for counting neighbors are consistent:

• $\mathrm{A-has-0}\lor \mathrm{A-has-1}$
• $\mathrm{A-has-0}\implies \neg \mathrm{A-has-1}$
• $\mathrm{A-has-1}\implies \neg \mathrm{A-has-0}$
• $\mathrm{B-has-0}\lor \mathrm{B-has-1}\lor \mathrm{B-has-2}$
• $\mathrm{B-has-0}\implies (\neg \mathrm{B-has-1}\land \neg \mathrm{B-has-2})$
• $\mathrm{B-has-1}\implies (\neg \mathrm{B-has-0}\land \neg \mathrm{B-has-2})$
• $\mathrm{B-has-2}\implies (\neg \mathrm{B-has-0}\land \neg \mathrm{B-has-1})$
• $\mathrm{H-has-0}\lor \mathrm{H-has-1}\lor \mathrm{H-has-2}\lor \mathrm{H-has-3}$
• $\mathrm{H-has-0}\implies (\neg \mathrm{H-has-1}\land \neg \mathrm{H-has-2}\land \neg \mathrm{H-has-3})$
• $\mathrm{H-has-1}\implies (\neg \mathrm{H-has-0}\land \neg \mathrm{H-has-2}\land \neg \mathrm{H-has-3})$
• $\mathrm{H-has-2}\implies (\neg \mathrm{H-has-0}\land \neg \mathrm{H-has-1}\land \neg \mathrm{H-has-3})$
• $\mathrm{H-has-3}\implies (\neg \mathrm{H-has-0}\land \neg \mathrm{H-has-1}\land \neg \mathrm{H-has-2})$

Axioms asserting that the safety propositions are consistent:

• $\mathrm{A-safe}\implies \neg \mathrm{A-unsafe}$ ,
• $\neg \mathrm{A-safe}\implies \mathrm{A-unsafe}$ ,
• $\mathrm{Z-safe}\implies \neg \mathrm{Z-unsafe}$ ,
• $\neg \mathrm{Z-safe}\implies \mathrm{Z-unsafe}$ .

This set of axioms is not quite complete, as explored in an exercise .

As mentioned, it is redundant to have both $\mathrm{A-safe}$ and $\mathrm{A-unsafe}$ as propositions. Furthermore, having both allows us to express inconsistent states(ones that would contradict the safety axioms). If implementing this in a program, you might use both as variables,but have a safety-check function to make sure that a given board representation is consistent.Even better, you could implement WaterWorld so that these propositions wouldn't be variables,but instead be calls to a lookup (accessor) functions. These would examine the same internal state,to eliminate the chance of inconsistent data.

Using only true/false propositions; without recourse to numbers makes these domain axioms unwieldy.Later, we'll see how relations and quantifiers help us model the game of WaterWorld more concisely.