On the difference between $\vdash A\to B$ and "If $\vdash A$, then $\vdash B$" I noticed that in general, the statements $\vdash A\to B$ and "If $\vdash A$ then $\vdash B$" are not equivalent.$^1$
However, this shows that I have a faulty intuition:

*

*I thought that I can interpret $\vdash A$ ($A$ is derivable without open assumptions) as "$A$ is true" without causing any harm.


*Accordingly, I interpret "If $\vdash A$ then $\vdash B$" as "If $A$ is true, then $B$ is true", which is the same as $\vdash A\to B$.
I could just accept that I have to be more careful, but I was hoping that someone could comment on this and give me some additional insight.

$^1$ For example, consider $B=\forall _x A$:
$$\text{If }\vdash A\text{, then }\vdash\forall_xA$$
is always true according to the rules of natural deduction, but of course
$$\vdash A\to\forall_xA$$
can only be derived if $x$ is not a free variable of $A$ (otherwise we could derive absurd formulas).
 A: 
I thought that I can interpret $\vdash A$ ($A$ is derivable without open assumptions) as "$A$ is true" without causing any harm.

Sorry, but no.  $\vdash A$ is a much stronger statement than simply "$A$ is true". As you point out, it means that $A$ can be derived without any assumptions, and that shows that $A$ isn't just True, but that $A$ is always true: $A$ is a tautology!
Consider the statement "It rains". This statement could be true or false, depending on where and when (basically, in what world) you evaluate it.
This is quite different from a statement like "it rains or it doesn't rain": that statement is true no matter what, and we call it a tautology.
In propositional logic, the first statement would be like $P$, and we cannot prove $P$ from no assumptions whatsovever. But the second statement is of the form $P \lor \neg P$, and that statement we can prove without any further assumptions. That is, where $P$ is an atomic statement, we have $\not \vdash P$, but we do have $\vdash P \lor \neg P$
If for your  statement $A$ it is the case that $\vdash A$, then $A$ is like the second statement, not the first.
