I am not a mathematician, though I am aware that:

  1. Any forall-statement about empty set is (vacuously) true because $\neg{(\forall x \in \{\}: P)} \rightarrow \exists x \in \{\}: \neg P$, where $\exists x \in \{\} \equiv False$ by definition: empty set.. is empty!
  2. Implication has kind of useless "special case" - $False \rightarrow True$ - when precondition is false and yet the consequence holds. Technically, this particular situation has nothing to do with if-else because it is still unknown whether $True \rightarrow True$ will hold as well. Never the less, $False \rightarrow True \equiv True$.

It seems to me that math is driven by the following philosophical principle:

Everything is true unless the opposite is proven.

In the #1 it is necessary to find such $x \in \{\}$ that ..., which is impossible. Being unable to prove "the opposite" implies undeniable truth. In the #2 it is necessary to show the case when precondition holds and consequence doesn't: unless it is shown, implication considered to be truthful.

Am I right?

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    $\begingroup$ The first one is a consequence of the second : $\forall x (x \in \emptyset \to P(x))$ is true because $\text F \to \text {whatever}$ is $\text T$. $\endgroup$ – Mauro ALLEGRANZA May 28 '19 at 9:37
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    $\begingroup$ I do not think "Everything is true unless the opposite is proven" is correct. Thanks to Gödel and Tarski we know there are true statements in mathematics which can be stated but cannot be proven within given systems; their negations would then be false statements whose opposite cannot be proven $\endgroup$ – Henry May 28 '19 at 9:38
  • $\begingroup$ While the conditional has that "unpleasant" feature ? Because we want that $A \to B$ is FALSE exactly when $A$ is TRUE and $B$ is FALSE. $\endgroup$ – Mauro ALLEGRANZA May 28 '19 at 9:39
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    $\begingroup$ See well-know Fermat's Last Th example : "Everything is a conjecture unless it is proved (or rejected providing a counterexample)". $\endgroup$ – Mauro ALLEGRANZA May 28 '19 at 9:40

I would rather state it as

Anything might be true until the opposite is proven.

If something is assumed to be true without a proof, someone else might say that the opposite is true, e.g. your claim is the opposite of his truth claim, so now you need to prove him wrong.

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  • $\begingroup$ Thought some time on your statement. Not quite agree: there is no "might be true" in #1 and #2 I've mentioned: both are definitely true, without any "might" or "may". What do you think about it? $\endgroup$ – Sereja Bogolubov May 28 '19 at 20:49

First, truth and provability are two different properties. To check if something is true or not you need a model. For example, if you talk about addition - you need some structure on which addition is defined - like in natural numbers sentence $\forall x \forall y \exists z: x = y + z \vee y = x + z$ is true, while $\forall x: x + x = 0$ is false.

To check if something is provable, you need axioms and deduction rules. For example, if you have axioms $x$ and $x \to y$ and MP rule: $\{a, a \to b\}\vdash b$, then you can deduce $y$.

For "$\forall x\in X\colon P(x)$ is true if $X = \varnothing$" you can consider the following game: you you can choose $x \in X$ s.t. $P(x)$ is false then you win, otherwise I win; statement $\forall x\in X\colon P(x)$ is true if I win in this game assuming you play optimally. Of course if $X = \varnothing$ you can't win.

For $\bot \rightarrow \top$ is true, take for example statement "if number is greater than $4$, it's greater than $2$". We probably want to think this statement is true (for any number). Lets check cases:

1) if the number is $5$, then it's of form $\top \rightarrow \top$

2) if the number is $1$, then it's of form $\bot \rightarrow \bot$

3) if the number is $3$, then it's of form $\bot \rightarrow \top$

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