An flawed proof about quantifiers

Question

Consider a propositional function $P(a)$ with a definite truth value for elements from $D$.

I am going to show that $\forall x \in D (P(x)) \equiv \exists x \in D (P(x))$

(obviously this is not true, but I would like to find the specific flaw in my reasoning)

I will be using these 4 logical equivalences:

Let $c$ be an arbitrary element from $D$.

A. $P(c) \equiv \forall x \in D (P(x))$

B. $P(c) \equiv \neg (\neg P(c))$

C. $\neg P(c) \equiv \forall x \in D (\neg P(x))$

D. $\neg (\forall x \in D (\neg P(x))) \equiv \exists x \in D (P(x))$

Proof: $\forall x \in D (P(x)) \equiv P(c) \equiv \neg (\neg P(c)) \equiv \neg (\forall x \in D (\neg P(x))) \equiv \exists x \in D (P(x))$

What have I done wrong here to reach this incorrect conclusion? Specifically, what rules of propositional logic have I violated?

Answer 1

Equivalences such as between $P(c)$ holding for arbitrary $c$ and $\forall x\,P(x)$ are not logical equivalences $P(c)\leftrightarrow \forall x\,P(x)$ that, like equality, would allow us to swap these two anywhere when they appear nested deeply in a complex formula. Rather, they are (in this case, bidirectional) rules of inference that are per se applicable only at "top level".

In other words, we have $$ P(c)\vdash \forall x\,P(x)$$and $$\forall x\,P(x)\vdash P(c), $$ but in general not $$ f(P(c))\vdash f(\forall x\,P(x))$$ and not$$ f(\forall x\,P(x))\vdash f(P(c)).$$ In particular, we do not have $$ \neg\forall x\, P(x)\vdash \neg P(c)$$ (though $$ \neg P(c)\vdash \neg\forall x\, P(x)$$ happens to be correct, but not by just a single inference step)

Answer 2

The problem is that your A. and C. are incorrect; the equivalence $P(c) \equiv \forall x P(x)$ does not hold (and neither with a negation). We have $\forall x P(x) \vDash P(c)$, but $P(c) \nvDash \forall x P(x)$.

$\forall x P(x) \vDash P(c)$ is an application of the rule of universal instantiation: $\forall x P(x) \vDash P(x)[x/t]$, for an arbitrary term $t$. If $P$ holds of all objects, then it holds of each instance of individuals.

There is no rule that would allow one to infer $P(c) \vDash \forall x P(x)$: From $P$ holding of one concrete object $c$ we can not infer that it holds for all objects.
There is the rule of universal generalization, but that works differently: It requires that 1. the term in the premise be a variable $y$, not a constant $c$, and that 2. the variable be arbitrary, in the sense that $P(y)$ does not depend on any open assumptions about $y$.
In $P(c)$, $c$ is not an arbitrary object, but rather $P(c)$ still depends on an open assumption, namely the assumption $P(c)$ itself, which is not proven.

You do have $\forall x P(x) \vDash P(c) \vDash \exists x P(x)$, but the reverse direction does not hold: Neither can you infer from the truth of $P$ for one particular object $c$ that $P$ holds of all objects.