Quantifiers and If then statements

Question

There are two sentences $\varphi(u), \psi(u)$ with common variable $u$. Are they equivalent: $\forall u[\varphi(u) \rightarrow \psi(u)]$ and $[\varphi(u) \rightarrow \psi(u)]$?

Answer 1

$\forall u[\varphi(u) \rightarrow \psi(u)]$ doesn't have any free variables. It's either true or false, depending on what $\varphi$ and $\psi$ say.

On the other hand, $[\varphi(u) \rightarrow \psi(u)]$ has a free variable (namely $u$). It doesn't have a set truth value untill you give it a concrete $u$ to evaluate, and then it is true or false only depending on that specific $u$.

This means that $\forall u[\varphi(u) \rightarrow \psi(u)]$ is true iff $[\varphi(u) \rightarrow \psi(u)]$ always evaluates to true, no matter what $u$ is; that is what $\forall$ means, after all.

Answer 2

By definition:
We say that formulas $\phi (x_1,x_2, \dots , x_n)$ and $\psi (x_1,x_2, \dots , x_n)$ are equivalent, if: $\forall_{x_1}\forall_{x_2} \dots \forall_{x_n} \ \phi (x_1,x_2, \dots , x_n) \leftrightarrow \psi (x_1,x_2, \dots , x_n)$ is a rule of logic. In other words two formulas are equivalent if theirs universal closures are equivalent. Applying it to our problem, we obtain: $$\forall_u [\forall_u[\phi (u) \rightarrow \psi(u)]] \leftrightarrow [\forall_u[\phi (u) \rightarrow \psi(u)],$$ which is obviously true.