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I came across an expression in predicate logic that made me wonder whether it was actually syntactically valid, and if so, semantically correct.

For a sentence like "Every dog chases a cat", there are two readings. (a) One where every dog there is, chases at least one cat and (b) one where at least one cat is chased by all dogs there are.

(a) $\forall x\in \{x : dog(x)\}:\ \exists y\in \{x : cat(x)\}:\ chase(x, y)$

(b) $\exists y\ \in\ \{x : cat(x)\}:\ \forall x\ \in\ \{x : dog(x)\} :\ chase(x,y)$

These two representations mean the same thing:

(a) $\forall x.[ dog(x) \rightarrow \exists y.[ cat(y) \land chases(x, y) ] ]$

(b) $\exists y.[ cat(y) \rightarrow \forall x.[ dog(x) \land chases(x, y) ] ]$

Now the expression I came across is this:

$$\forall x.(dog(x) \rightarrow \exists y.(cat(y) \land chase(x))(y))$$

This looks weird to me. To make it more apparent let more transform it:

It is the same as:

$$\forall x.[ dog(x) \rightarrow \exists y.[ cat(y) \land chase(x) ] (y) ]$$

which is the same as:

$$\forall x.[ dog(x) \rightarrow \exists y.[ cat(y) \land chase(x, ] y) ]$$

First of all, is it syntactically valid to have "quantifier[ functor( ] )"? So, having overlapping projections of syntactic nodes in the syntactic tree of the expression?

And if the expression is syntactically valid... what does it mean? It doesn't make any sense to me.

user3578468
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1 Answers1

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Before answering your question, I must cover some aspects of a formal first-order language.

Particularly, the definition of its syntax. From what I stated previously here and here:

I assume you are referring to a first-order language $L_\Sigma=<\Sigma,V,\neg,\rightarrow,\forall>$, where

  • $\Sigma$ is a triple $<R,F,C>$ the signature of $L$, where $R=\{R_n\}_{n \in \mathbb{N}}$ and $F=\{F_n\}_{n \in \mathbb{N}}$ are family of sets and $C$ is a set. The elements of $R$ are called $n$-ary relation symbols, the elements of $F$ are called $n$-ary function symbols, the elements of $C$ are called constant symbols. We say that $\Sigma$ is the signature of $L$.
  • $V$ is a set, called the variable symbols of $L_\Sigma$.

Now let $L_\Sigma^*$ denote the set of all strings of $L_\Sigma$.

The usual definition of the set of well-formed formula of $L_\Sigma$, call it $FORM_{L_\Sigma}$ is then:

Definition (FORM) The set FORM of well-formed formulas of $L_\Sigma$ is the set satisfying:

  1. $R(a_1,...,a_n) \in FORM$ if $a_1,...,a_n \in TERM$
  2. $(\neg\alpha) \in FORM$ if $\alpha$ is a wff
  3. $(\alpha \rightarrow \beta) \in FORM$ if $\alpha$ and $\beta$ are wff
  4. $(\forall v (\alpha)) \in FORM$ if $v \in V$ and $\alpha$ is a wff
  5. No string that is not obtained by (1),(2),(3) or (4) is in FORM.

Two observations:

  • I assume the definition of the set of terms of $L_\Sigma$, $TERM$. You may try as an exercise.

  • The existential quantifier is defined as $$ \exists x \alpha \equiv \neg \forall x \neg \alpha$$

Now, look at (1), and (4). Does this definition allow us to get the sentences you asked?

  • Thanks. Uhm, to me it looks like according to your definition the expression in question (the last one with the intersecting brackets) is not a member of FORM... but... I don't know if that makes it universally invalid as a predicate logic expression. – user3578468 May 31 '15 at 04:18
  • Yes, it's not a member of $FORM$, therefore not a well-formed formula. – Bruno Bentzen May 31 '15 at 04:26
  • But certainly there are languages where crossing edges of projections are valid...? But the set of valid expressions in predicate logic is not one of those languages? – user3578468 May 31 '15 at 04:28
  • @user3578468 I can't see any practical reason for admitting such kind of syntax - besides, think about how would you define the semantics of this "crossing edges of projections" syntax you propose? – Bruno Bentzen May 31 '15 at 04:37
  • I don't. I just considered the option, that the expression might indeed be semantically interpretable with a meaning that didn't reveal itself to me. All under the assumption that it might be syntactically permissible, contrary to my intuition of it being not. But thanks a lot for helping me out! – user3578468 May 31 '15 at 04:49
  • @user3578468 this is simply a miswritten expression. Perhaps a typo. That's all. – magma May 31 '15 at 20:03
  • @magma Right to the point :) – Bruno Bentzen Jun 01 '15 at 00:59