How to interpret $\forall _{x: \phi(x)} \psi(x)$ versus $\forall _x (\phi(x) \wedge \psi (x))$

Question

It's clear to me that $$ (\forall _{x: \; \phi(x)} \;\psi(x)) \equiv (\forall_x \;\phi(x) \; \rightarrow \; \psi(x)).$$

I would like to know the relationship between these two expressions: $$ \forall _{x: \; \phi(x)} \;\psi(x)) \;\; \text{versus} \;\; \forall _x (\phi (x) \wedge \psi(x))$$

Does one imply the other?

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EDIT

If $\phi (x) $ is "x is prime-and-bigger-than-2" and $\psi (x)$ is "x is odd", then $\forall _{x: \phi(x)} \psi(x)$ is "Every number which is prime-and-bigger-than-2 is odd".

I think the main difficulty in interpreting is comparing "... which X is Y" and "... is both X and Y".

Answer 1

The relation is that $\forall x \mathop. \varphi(x) \land \psi(x)$ implies $\forall x \mathop. \varphi(x) \to \psi(x)$ but the converse does not hold.

We can prove that the converse does not hold by considering the following assignment of concrete expressions to metavars: $\varphi(\cdots) = \bot$ and $\psi(\cdots) = \top$.

We can prove the implication by assuming $\forall x \mathop. \varphi(x) \land \psi(x)$ and $\lnot \forall x \mathop. \varphi(x) \to \psi(x)$ and deriving a contradiction. This is not a proof in a formal proof calculus like a tableau calculus, or system LK, or a Fitch-style proof, or a Hilbert system.

• premise: $\forall x \mathop. \varphi(x) \land \psi(x)$
• negated goal: $\lnot \forall x \mathop. \varphi(x) \to \psi(x)$
• $\exists a \mathop. \varphi(a) \land \lnot \psi(a)$
• We can introduce a fresh constant: $\varphi(c) \land \lnot \psi(c)$
• Thus $\lnot \psi(c)$.
• From our premise, we conclude that $\varphi(c) \land \psi(c)$
• And thus $\psi(c)$
• Contradiction.

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So, Alonzo Church's book Introduction to Mathematical Logic has a very nice notation $\varphi(x) \supset_x \psi(x)$ (introduced on page 44), which is equivalent to $\forall x \mathop. \varphi(x) \to \psi(x)$ in modern notation. Sometimes the combination of a quantifier and the toplevel connective of the clause inside it collectively have a nice interpretation in natural language.

I'll borrow this notation and let $\varphi(x) \to^x \psi(x)$ be equivalent to $\forall x \mathop. \varphi(x) \to \psi(x)$ and likewise for $\land$.

The first, $\varphi(x) \to^x \psi(x)$, is equivalent in prose to all $x$s that are $\varphi$ are $\psi$.

The second, $\varphi(x) \land^x \psi(x)$, is equivalent in prose to all $x$s are both $\varphi$ and $\psi$.

When phrased that way, it is clear that the second is a much stronger statement than the first and that the second implies the first.

Equivalently, the first is saying that the set of all $\varphi$s is a subset of the set of all $\psi$s and the second is saying everything in the universe is a $\varphi$ and everything in the universe is a $\psi$.