weird natural deduction question

Question

There are these two questions that my professor posted, and they absolutely stumped me:

$ \vdash (\exists x. \bot) \implies P $ and $(\exists x. \top) \vdash (\forall x. \bot ) \implies P$.

What do I even do with the $(\exists x. \bot)$ part? It got me stuck for quite some time. Any help will be appreciated.

However, I wonder why the second problem doesnÄt simply read $\vdash (\forall x.\bot)\to P$; after all $\forall$-elimination allows $(\forall x.\bot)\vdash \bot$. — Hagen von Eitzen, Feb 13 '18 at 18:54
@HagenvonEitzen Not if you allow an empty domain which the assumption of $\exists x.\top$ rules out. — Derek Elkins left SE, Feb 13 '18 at 19:58
This looks like the kind of problem that would be proposed if you were studying inclusive logics (that is logics which don't presuppose that the universe of discourse is nonempty). Do you have a reference for the logic your professor is using? — DanielV, Feb 14 '18 at 02:53

score 3 · Accepted Answer · answered Feb 13 '18 at 19:00

3

In a syntax that allows vacuous quantification (bad and un-Fregean, but now sadly usually permitted), $\exists x\,\phi$ is equivalent to plain $\phi$, where $\phi$ is closed, without free variables, and so $\exists x\,\bot$ is equivalent to plain $\bot$. And you know, presumably, about the ex falso principle $\vdash \bot \to P$. Put those two together to get the first result.

Can you now see why the second result similarly holds?

answered Feb 13 '18 at 19:00

Peter Smith

54,743

.. dammit. so basically the first one means that there exists some $x$ in the empty set, which means vacuously true, so any P will hold base on false assumption.
So if I am to do natural deduction, $(\exists x. \bot)$ simply leads to $\bot$, then i can just proceed freely?

for second one, $(\exists x .\top)$ simply leads to true statement?
– Kenny Wong Feb 13 '18 at 19:03
No. The official definition of $\exists x \varphi$ tells you that this is true if and only if there exists some $x$ in the relevant domain which satisfies $\varphi$. Now work out from the official definition of satisfaction what this comes to in the special case where $\varphi$ is a closed sentence, that doesn't contain $x$ free. – Peter Smith Feb 13 '18 at 20:02
hmm ok, but how do i put it if i am doing natural deduction? i mean, $(\exists x. \bot)$ will have to lead to $\bot$ somehow right? what will be the reasoning behind it? – Kenny Wong Feb 13 '18 at 20:11

score 1 · Answer 2 · answered Feb 13 '18 at 21:10

Here are some proofs in the Fitch system:

$\def\fitch#1#2{\begin{array}{|l}#1 \\ \hline #2\end{array}}$

$\fitch{ 1. }{ \fitch{ 2.\exists x. \bot}{ \fitch{ 3.\bot }{ 4.P \quad \bot \text{ Elim } 3} \\ 5.P \quad \exists \text{ Elim } 2, 3-4} \\ 6. \exists x. \bot \rightarrow P \quad \rightarrow \text{ Intro } 2-5}$

$\fitch{ 1. \exists x. \top }{ \fitch{ 2.\forall x. \bot}{ 3.\bot \quad \forall \text{ Elim } 2\\ 4.P \quad \bot \text{ Elim } 3} \\ 5. \forall x. \bot \rightarrow P \quad \rightarrow \text{ Intro } 2-4}$

Note that for the second proof you never use the $\exists x. \top$. Indeed, $\forall x. \bot \rightarrow P$ is valid all by itself.

weird natural deduction question

2 Answers2