Do ∃x(Dog(x)) and ∃x(¬Dog(x)) contradict each other?

Question

Formally, ∃x(Dog(x)) and ∃x(¬Dog(x)) look like they contradict each other. However, in the real world,

1. there exist objects which are dogs i.e. ∃x(Dog(x))
2. there exist objects which are not dogs i.e. ∃x(¬Dog(x))

I don't know what I'm getting wrong. I was inspired to ask this question after watching a YouTube video on translating "Not all dogs are happy" to predicate logic. I initially obtained ∃x(Dog(x)∧¬Happy(x)): "there exists an object which is a dog and it's not happy". I'm not sure if this translation is correct either.

Answer 1

in the real world,

1. there exist objects which are dogs i.e. ∃x(Dog(x))

In other words: "something is a dog".

2. there exist objects which are not dogs i.e. ∃x(¬Dog(x))

In other words: "not everything is a dog", i.e., ¬∀x(Dog(x)).

Formally, ∃x(Dog(x)) and ∃x(¬Dog(x)) look like they contradict each other.

As you have demonstrated above, $(1)$ and $(2)$ are in fact consistent with each other; so, they certainly do not contradict each other.

To be clear: the negation of $(2)$ is ∀x(Dog(x)) rather than $(1)$.

Answer 2

A contradiction is a well-formed formula of the form

$$P \wedge \neg P $$

So, if we know $P$ and want to derive a statement that contradicts $P$, then we simply need to determine $\neg P$. With this in mind, if we let the proposition $P : \exists x Dx$ where $Dx : $ "$x$ is a dog" and the domain is the set of all things, then

$ \begin{array}{llll} \neg P \\ \neg \exists x Dx & \text{by definition of $P$} \\ \forall x \neg Dx & \text{by DeMorgan's Law for Quantifiers} \\ \end{array} $

Hence, the formulas $\exists x Dx$ and $\forall x \neg Dx$ contradict each other. This becomes more apparent when we consider their translations

$ \begin{array}{llll} \exists x Dx : & \text{There exists at least one $x$ such that $x$ is a dog.} \\ \forall x \neg Dx : & \text{For every $x$, it's not the case that $x$ is a dog.} \end{array} $

In short, we're saying "there exists a dog" and "there are no dogs" in the domain, respectively. Obviously, these two formulas are inconsistent, so a conjunction of the two will always be false.

Your translation of "not all dogs are happy" is correct. First, notice the statement is actually the negation of a universal statement stating "all dogs are happy." Given

$ \begin{array}{llll} Dx: & \text{$x$ is a dog} \\ Hx: & \text{$x$ is happy} \end{array} $

we translate "all dogs are happy" as $\forall x [Dx \to Hx]$. In other words, "for every $x$, if $x$ is a dog, then $x$ is happy." The negation of this formula is as follows

$ \begin{array}{llll} \neg \forall x [Dx \to Hx] \\ \exists x \neg [Dx \to Hx] & \text{DeMorgan's Law for Quantifiers} \\ \exists x \neg [\neg Dx \vee Hx] & \text{Implication Rule} \\ \exists x [\neg\neg Dx \wedge \neg Hx] & \text{DeMorgan's Law} \\ \exists x [Dx \wedge \neg Hx] & \text{Double Negation Elimination} \\ \end{array} $

which reads as "there exists at least one $x$ such that $x$ is a dog and $x$ is not happy."