Does this means that "anything" implies $T$? I'm reading Yves Nievergelt's Logic, Mathematics and Computer Science. Here: 



I am very confused about this. I understand the proof, but does that mean that anything implies $T$? Supposing my interpretation is correct, isn't it weird that anything implies $T$?  
 A: Semantically, to assert $S\Rightarrow T$ is the same as asserting that it is impossible for $S$ to be true and at the same time $T$ is false.
When $T$ is a theorem, it means that we know that $T$ is actually true. So the only situation that could prevent $S\Rightarrow T$ from being true is impossible right out of the gate -- we don't even need to look at $S$. Therefore $S\Rightarrow T$ cannot be false, so it must be true, and therefore it ought to be a theorem.
A: The hypothesis is that $T$ is a theorem ("for each well-formed formula $S$ and each theorem $T$"). This means that we have a proof of $T$ from nothing at all (that is, $\vdash T$). Adding hypotheses doesn't make things harder, so that same proof is a proof of $T$ from $S$, whatever $S$ is.
So yes, a $T$ of the type described in the theorem is provable from any $S$ whatsoever. Put another way, if $T$ is a theorem then $(S)\implies (T)$ is also a theorem for any (well-formed) $S$.
A: 
I am very confused about this. I understand the proof, but does that mean that anything implies T? Supposing my interpretation is correct, isn't it weird that anything implies T?

It is not weird.   If $T$ is indeed true, then surely that anything should imply that it is true.
$S\to T$ can be interpreted as : $T$ is true whenever $S$ is true.   That should be true when $T$ be guaranteed to be true, whatever $S$ may be.
