The domain in the First Order Logic expression

discrete mathematicsfirst-order-logic

I think I'm misunderstanding how to apply the domain for these expressions.

The question is:

Let $P(x), Q(x), R(x),$ and $S(x)$ be the statements “$x$ is a duck,” “$x$ is one of my poultry,” “$x$ is an officer,” and “$x$ is willing to waltz,” respectively. Express each of these statements using quantifiers; logical connectives; and $P(x), Q(x), R(x),$ and $S(x)$.

One of my answers.

  1. No ducks are willing to waltz.

$∃x(S(x) \to ¬P(x))$

The answer that pops up when I google.

$∀x(S(x) \to ¬P(x))$

The way that I interpret this, the domain is everyone and my answer reads "If somebody is a duck, then they are not willing to waltz".

I interpret the seemingly correct answer as, "Everyone is a duck, so they are not willing to waltz".

Best Answer

Your issue isn't with domains, it's with how to parse the logical expressions syntactically.

Look at the parentheses, and be careful to translate quantifiers very literally. An expression of the form "$\exists x(stuff)$" means "There is some $x$ such that $stuff$." Similarly, an expression of the form "$\forall x(stuff)$" means "For every $x$, $stuff$." So your proposed answer reads, in somewhat-natural language, as follows:

There is some $x$ such that if $x$ is willing to waltz then $x$ is not a duck,

or a bit more cleanly, as:

There is something which, if it is willing to waltz, is not a duck.

This is very obviously not the same as "No ducks are willing to waltz" - it merely indicates that there is at least one thing which is not a willing-to-waltz duck.

By contrast, the correct answer $\forall x(P(x)\rightarrow\neg S(x))$ translates to

For every $x$, if $x$ is a duck then $x$ isn't willing to waltz

which is indeed the same as

No duck is willing to waltz

if a bit more indirect sounding.

(As an aside, your incorrect reading of the correct answer as "Everyone is a duck, so they are not willing to waltz" also suggests another issue, namely that you're trying to read $\rightarrow$ as indicating causality somehow - per the word "so" in your reading.)

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