Beginner Predicate Logic confusion

Question

Learning predicate logic and I am coming across the following questions and I am somewhat confused as the what is expected in the answers, and when it is recommended to quantify secondary variables.

First, 'Some raccoons are chased by every dog in the town' I have transcribed as thus: Domain of y is the animals in the neighbourhood. R = is a Raccoon D = is a Dog C = chases Raccoons (∃x)(Dx ⊃ (∀y)(Cy & Ry)) Should I have quantified the second variable? Could I just have done 'There exists x, where x is a raccoon and is chased by cats?'

Secondly, 'If there are any peacocks that are blue, then some peacocks are male' I have translated to: B = is blue P = is a peacock M = is male (∃x)((Px & Bx) ⊃ ((∃x)( Px & Mx)) Should I have quantified a second variable for the second object, can you quantify the existence of a same variable? Thanks.

Answer 1

"Some raccoons are chased by every dog in the town."

Domain: $\{ x : x $ is a thing$\}$

$ \begin{array}{ll} Rx: & \text{$x$ is a raccoon.} \\ Dx: & \text{$x$ is a dog.} \\ Tx: & \text{$x$ is in the town.} \\ Cxy: & \text{$x$ chases $y$.} \\ \end{array} $

$ \exists x [Rx \wedge \forall y [Dy \wedge Ty \to Cyx]] \equiv \exists x \forall y [Rx \wedge [Dy \wedge Ty \to Cyx]]$

In other words, "There exists at least one $x$ such that $x$ is a raccoon and, for every $y$, if $y$ is a dog in the town, then $y$ chases $x$."

"If there are any peacocks that are blue, then some peacocks are male."

Domain: $\{ x : x $ is a thing$\}$

$ \begin{array}{ll} Bx: & \text{$x$ is blue.} \\ Px: & \text{$x$ is a peacock.} \\ Mx: & \text{$x$ is a male.} \\ \end{array} $

$ \exists x [Px \wedge Bx] \to \exists x [Px \wedge Mx]$

In other words, "If there exists at least one $x$ such that $x$ is a blue peacock, then there exists at least one $x$ such that $x$ is a male peacock."

Answer 2

First, 'Some raccoons are chased by every dog in the town' I have transcribed as thus: Domain of y is the animals in the neighbourhood. R = is a Raccoon D = is a Dog C = chases Raccoons (∃x)(Dx ⊃ (∀y)(Cy & Ry))

There are several things wrong with this translation.

First, "C = chases raccoons" is not a fine-grained enough decomposition of words for a single predicate as it contains both the concept of chasing and the raccoons which already have their own predicate; "C" should just be "chases" and take two arguments, the first for the chaser and the second one for the chased. In your formula, with $x$ being used to quantify over the raccoons and $y$ for the dogs, you get $Cyx$ for the dogs chasing the raccoons.

In your formula, you correctly start with an existential quantifier but then make it about the dogs; the subject of the existential quantifier should be the raccoons and the one of the universal quantifier the dogs: ($\exists x) (Rx ... (\forall y) (Dy ...))$.

Then you're using the wrong connectives with each quantifiers: Generally, $\exists$ goes with $\&$ and $\forall$ goes with $\supset$. So it should be $(\exists x) (... \& ... (\forall y) (... \supset ... ))$.

Plugging this together, the formula becomes

$$(\exists x) (Rx \& (\forall y) (Dy \supset Cyx))$$

Should I have quantified the second variable?

Not sure what you mean here; both variables are already quantified in your formula, $x$ by $\exists$ and $y$ by $\forall$.

Could I just have done 'There exists x, where x is a raccoon and is chased by cats?'

No, your idea of breaking it down some more into a predicate for the dogs and one for the chasing and an explicit quantifier for the "every dogs" was correct. Just shoving all that into a single predicate would be a rather imprecise formalization.

Secondly, 'If there are any peacocks that are blue, then some peacocks are male' I have translated to: B = is blue P = is a peacock M = is male (∃x)((Px & Bx) ⊃ ((∃x)( Px & Mx)) Should I have quantified a second variable for the second object, can you quantify the existence of a same variable? Thanks.

This one is fine. You could use a different variable for clarity, but since the scope of each existential quantifier is limited by the brackets around the &-ed statement, they are technically separated.