Does this means that "anything" implies $T$?

Question

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 $T$?

Well, what would be an example of something you think shouldn't imply $T$? — Eric Wofsey, May 15 '19 at 21:22
Not quite this question, but potentially enlightening: In classical logic, why is $p\Rightarrow q$ True if $p$ is False and $q$ is True? — hmakholm left over Monica, May 15 '19 at 21:24

score 6 · Answer 1 · answered May 15 '19 at 21:30

6

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.

answered May 15 '19 at 21:30

hmakholm left over Monica

286,031

But does that mean that we could find a proof that $S\implies T$, for any $S$? – Red Banana May 15 '19 at 21:42
@BillyRubina Yes, if $T$ is provable, then $S\to T$ is too, whatever $S$ may be. – Graham Kemp May 15 '19 at 22:46
@BillyRubina: Yes, and the argument you show in the question tells you exactly how to construct that proof. – hmakholm left over Monica May 16 '19 at 00:03

score 2 · Answer 2 · answered May 15 '19 at 22:29

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) $S$.

score 2 · Answer 3 · answered May 16 '19 at 01:08

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.

Does this means that "anything" implies $T$?

3 Answers3