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I am having a hard time distinguishing between the two different logics. If we consider the statement, “Squares adjacent to the Wumpus are smelly,” and are asked to express it as First-Order Logic, we can express First-Order Logic as so:

\begin{gather} \forall x,y.\left(\left(\mathit{Square}(x) \land \mathit{Wumpus}(y) \land \mathit{Adjacent}(x,y)\right) \to \mathit{Smelly}(x)\right) \\ \forall x,y,a,b.\mathit{Adjacent}([x, y],[a, b]) \leftrightarrow (x= a \land (y=b-1 \lor (y=b+1)) ) \lor (y=b \land (x=a-1 \lor (x=a+1)) ) \end{gather}

But I have friends who expressed their answers as so:

$$ \forall x,y. \mathit{Wumpus}([x, y]) \leftrightarrow \mathit{Smelly}([x - 1, y]) \land Smelly([x + 1 , y]) \land \mathit{Smelly}([x , y - 1]) \land \mathit{Smelly}([x , y + 1]) $$

My argument that their answers are not First-Order Logic, but rather Propositional Logic because there is a need to repetitively write $\mathit{Smelly}$ for all tiles that are smelly. I think my argument might be a little vague.

So, is the answer provided by my friend correct and why?

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  • $\begingroup$ Maybe it's because I'm not 'merrican, but the fact that you use these abstruse examples doesn't help forming an answer. All I see is that you seem to imply a logic which knows of arithmetic, because a there is appearently addition defined over Wumpi. And some sort of bracket. Anyhow, if there are universal quantiers, then it's hardly a proposition formed from true/false statements. $\endgroup$ – Nikolaj-K Jun 12 '13 at 8:55
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    $\begingroup$ Your friend(s)'(s) formulation is in first-order logic, and it would be correct if the $\iff$ were changed to $\implies$. I've never heard of proportional logic, so I'll guess that you meant propositional logic. Repetitions do not make a first-order sentence become propositional. The essential characteristic of propositional logic is that statements are built from propositional variables (not any other sort of variables) by means of propositional connectives (not quantifiers, not equality, not predicate symbols, and not function symbols). $\endgroup$ – Andreas Blass Jun 12 '13 at 14:20
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Both formalizations of the sentence are in first-order logic. The propositional calculus does not contain quantification, so something starting with $\forall x,y \dots$ isn't a propositional formula.

However, you have hit upon an interesting point. When the domain of discourse is known to be finite, e.g., the set $\{a,b,c\}$, then we can replace the formulae

$$ \forall x.\phi(x) \qquad \exists x.\phi(x)$$

with the equivalent formulae

$$ \phi(a) \land \phi(b) \land \phi(c) \qquad \phi(a) \lor \phi(b) \lor \phi(c)$$

When the domain is fixed, you can use this technique to remove quantifiers from all your formulae. For instance, in a smaller domain, $\{a,b\}$, the sentence

$$ \forall x.\exists y.P(x,y) $$

can be replaced first by

$$ \exists y.P(a,y) \land \exists y.P(b,y) $$

which in turn is replaced by

$$ (P(a,a) \lor P(a,b)) \land (P(b,a) \lor P(b,b)) $$

Here we have an opportunity to replace some sentences in first-order logic with sentences in the propositional calculus. We can replace each first-order ground literal with a propositional symbol, and turn the previous sentence into

$$ (P_{a,a} \lor P_{a,b}) \land (P_{b,a} \lor P_{b,b}) $$

which is in the propositional calculus.

To propositionalize a Wumpus world, you would take your grid (for ease, let's assume it's 3×3), and identify each cell in the grid:

\begin{array}{|c|c|c|} \hline A & B & C \\ \hline D & E & F \\ \hline G & H & I \\ \hline \end{array}

Now, instead of the arithmetic based definition of adjacent, you simply have the long propositional formula:

\begin{gather} \lnot A_{A,A} \land A_{A,B} \land \lnot A_{A,C} \land A_{A,D} \land \lnot A_{A,E} \land \dots \land \\ A_{B,A} \land \lnot A_{B,B} \land A_{B,C} \land \lnot A_{B,D} \land A_{B,E} \land \dots \land \\ \dots \end{gather}

You can keep applying this process to your other definitions, and entirely propositionalize the formalization of the domain. The number of sentences will be much bigger, but there are very efficient propositional reasoners, so in some cases the translation can be justified.

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  • $\begingroup$ Two comments about this: (1) The question seems to allow for the possibility that the Wumpus world is infinite. Indeed, the use of expressions like $a+1$ and $a-1$ strongly suggests that it is $\mathbb Z^2$. (2) Even if the world is intended to be finite, the conversion from first-order to propositional form works only if you know which finite universe is intended. $\endgroup$ – Andreas Blass Jun 12 '13 at 22:06
  • $\begingroup$ @AndreasBlass That's right; propositionalization really only works well when the domain is finite and you know something about it. One of the things I thought about pointing out about each of the first-order sentences in the question is that they might not handle the edges (if the world is a grid) very well. $\endgroup$ – Joshua Taylor Jun 12 '13 at 22:20
  • $\begingroup$ Since propositional calculus is decideable, this implies that for a fixed finite domain, first-order formulas are decideable. $\endgroup$ – Doug Spoonwood Jul 22 '13 at 1:59

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