1
  1. $\;\bot$ --- premise
  2. $\;\varphi\vee\bot$ --- $\vee$ intro 1
  3. $\;\bot$ --- repeat 1
  4. $\;\bot\rightarrow\bot$, i.e. $\neg\bot$ --- $\rightarrow$ intro 3
  5. $\;\varphi$ --- MTP 2,4

Quote from Mauro Allegranza who suggested this derivation in a comment in this question:

Regarding $\rightarrow$-intro, I'm using the case: $A\vdash$ $B\rightarrow A$.

This is perfectly sound: if A is (assumed to be) true, then $B\rightarrow A$ is also true, under assumption A; see von Plato, page 33.

This "intuition" can be formalized through the fact that we can clearly derive A from assumption A. But we can always add "unnecessary" assumptions; thus, we have also $A,B\vdash A$ and so, by $\rightarrow$-intro: $A\vdash B\rightarrow A$.

It seems to me we can remove step 3 and have this:

  1. $\;\bot$ --- premise
  2. $\;\varphi\vee\bot$ --- $\vee$ intro 1
  3. $\;\bot\rightarrow\bot$, i.e. $\neg\bot$ --- $\rightarrow$ intro 1
  4. $\;\varphi$ --- MTP 2,3

Now what's your thought on this?

Pooria
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1 Answers1

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Well, as always, much depends on exactly what rules you have and how they are defined. For example:

  1. $\;\bot\rightarrow\bot$, i.e. $\neg\bot$ --- $\rightarrow$ intro 3

would not be acceptable in most proof systems, as you'd need to actually derive $\neg \bot$ from $\bot \rightarrow \bot$ ... but I assume that in this proof $\neg \varphi$ is defined as $\varphi \rightarrow \bot$

Also, typically (again, it depends on how the rule is typically defined!) $\rightarrow$ Intro requires an assumption that gets discharged. Inded, many systems use a subproof structure for this:

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

$\fitch{ 1. \bot \quad \text{Premise}} {\fitch{ 2. \bot \quad \text{Assumption}}{ 3. \bot \quad \text{Reiteration}} \\ 4. \bot \rightarrow \bot \quad \rightarrow \text{ Intro } 2-3}$

Since you have $\bot$ as a premise, rather than an assumption, both the proof by Mauro and your simplification would not be acceptable by this kind of proof system.

Now, you are right that the Reiteration step is not always necessary (again, it depends on exactly how the rule is defined!), and that this part of the proof can be simplified to:

$\fitch{ 1. \bot \quad \text{Premise}} {\fitch{ 2. \bot \quad \text{Assumption}}{ } \\ 3. \bot \rightarrow \bot \quad \rightarrow \text{ Intro } 2-2}$

Interestingly many proof systems can use the subproof to apply $\neg$ Intro, giving you the desired $\neg \bot$ explicitly:

$\fitch{ 1. \bot \quad \text{Premise}} {\fitch{ 2. \bot \quad \text{Assumption}}{ 3. \bot \quad \text{Reiteration}} \\ 4. \neg \bot \quad \neg \text{ Intro } 2-3}$

... and without Reiteration:

$\fitch{ 1. \bot \quad \text{Premise}} {\fitch{ 2. \bot \quad \text{Assumption}}{ } \\ 3. \neg \bot \quad \neg \text{ Intro } 2-2}$

Completing the last proof to derive $\varphi$ from $\bot$:

$\fitch{ 1. \bot \quad \text{Premise}} {\fitch{ 2. \bot \quad \text{Assumption}}{ } \\ 3. \neg \bot \quad \neg \text{ Intro } 2-2\\ 4. \bot \lor \varphi \quad \lor \text{ Intro } 1\\ 5. \varphi \quad \text{ MTP } 3,4}$

Bram28
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  • Now which one should we take as primitive and then derive the other one from it? MTP or $\bot\vdash\varphi$? depends on system again? – Pooria Sep 28 '17 at 15:59
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    @Pooria You got it! Of course, some systems have both .. and some systems have neither. :) – Bram28 Sep 28 '17 at 16:01
  • God it's complicated :D , btw am I wrong to assume if in a system $\bot\vdash\varphi$ was a primitive, the inventor of that system has necessarily derived it this way "if $\bot$ is true then $\bot$ or $\varphi$ is true and since $\neg\bot$ is true so $\varphi$ is true"? I think I'd better not go that far :) – Pooria Sep 28 '17 at 16:22
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    @Pooria nothing as complicated as that ... we have $\bot \vDash \varphi$ because it is impossible for $\bot$ to be true and $\varphi$ to be false at the same time. And since $\bot \vDash \varphi$ (that is, since by our semantics we find that any $\varphi$ is a logical consequence of $\bot$, we have a justification to define $\bot \vdash \varphi$ as a sound inference rule. – Bram28 Sep 28 '17 at 17:06
  • Hmmm... I currently don't know what $\vDash$ means, so I guess whole confusion is due to my lack of knowledge, there should be some stuff to study before jumping into natural deduction, thanks for making me aware, what book do you recommend ^^ ? – Pooria Sep 28 '17 at 17:23
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    @Pooria You know, I was thinking the same thing: when you were asking all these questions about what can be derived from what, or what must be assumed in order for something else to hold, you were couching them in terms of formal proof systems and natural deduction (which is about the $\vdash$), but most likely you were really asking about the underlying semantics of logic all along (which is about the $\vDash$). So yes, it would be really good for you to understand the difference. Eh ... you could just search for 'formal syntax' and 'formal semantics' of logic. – Bram28 Sep 28 '17 at 19:32
  • Hey legendary Bram ^^ , in your comment above you said "... it is impossible for $\bot$ to be true and $\varphi$ to be false at the same time.", is it $\bot\vDash\varphi$ in other words (which I have to study formal semantics to understand)? – Pooria Oct 05 '17 at 11:54
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    @Pooria Exactly! In general, where $\Gamma$ is a set of sentences, and $\varphi$ is a sentence, you have $\Gamma \vDash \varphi$ (" $\varphi$ is a logical consequence of $\Gamma$") if and only if it is impossible for all sentences in $\Gamma$ to be true while $\varphi$ is false all at the same time. – Bram28 Oct 05 '17 at 12:33
  • And so then in formal semantics It's possible to assume $\bot$ is true? – Pooria Oct 05 '17 at 15:47
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    @Pooria Sure! You can always assume anything you want ... but in that particular case you'll have an immediate contradiction! :) – Bram28 Oct 05 '17 at 17:43
  • hmmm... wanted to delete that nonsense question but had not access to my PC :D (dunno if you've noticed but sometimes I ask something and delete it a few minutes later X'D ), btw can you point me to somewhere that explains how we can derive $\bot\vDash\varphi$? – Pooria Oct 05 '17 at 17:56
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    @Pooria Boy, I wish I could point you to a straightforward book, but I don;t know of one. However, typically formal semantics would say: $\bot$ always evaluates to false. Therefore, it is impossible for $\bot$ to be true and $\varphi$ to be false. Therefore, $\bot \vDash \varphi$. So this is not formal derivation (it's not a logic proof), but it's a mathematical ('normal') proof that simply refers to the formal semantics of a logical language that defines the mapping from expressions to truth-values. – Bram28 Oct 05 '17 at 19:50
  • hmmm I actually came up with this reasoning myself and asked about it but deleted it afterwards :D , so even formal (logical) proofs for $\bot\vDash\varphi$ are basically stating this reasoning? – Pooria Oct 06 '17 at 16:35
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    @Pooria No, formal proofs simply have to follow a pre-defined pattern ... that's the big difference between formal proofs (purely syntactical objects using purely syntactical rules) and proofs using formal semantics. So if you want a formal proof (i.e. if you want to show $\bot \vdash \varphi$, which you read as "there exists a formal proof from $\bot$ to $\varphi$), you either have a rule that says "when you have a $\bot$ you can write down anything you want", or you have to take multiple steps with other formally defined rules. – Bram28 Oct 06 '17 at 16:49
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    Here is another way to illustrate the difference: Suppose I define a formal proof system that has the following rule "At any point, you can write down any statement you want!" ... we could represent that rule as $\vdash \varphi$. Now, with that rule, I can prove $\vdash P$, or even $\vdash P \lor \neg P$, even though it is clear that $\not \vDash P$ and $\not \vDash P \lor \neg P$ . As such, we say that this formal rule is not sound ... but it is still a rule .. and hence we can still infer anything we want! - I call this the Hokus Ponens rule! :) – Bram28 Oct 06 '17 at 16:55
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    So that means there's no formal proof for $\bot\vDash\varphi$, we can only present a proof for it using formal semantics, correct? – Pooria Oct 06 '17 at 16:58
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    @Pooria Exactly!! Although if you know that your proof system is sound, then whenever we have $\Gamma \vdash \varphi$, we have $\Gamma \vDash \varphi$. (and of course, typical proof system in textbooks are sound ... ) But you're absolutely right, a formal proof of $\vdash \varphi$ does not by tiself imply $\vDash \varphi$ – Bram28 Oct 06 '17 at 16:59
  • Thanks for being a good teacher ^^ – Pooria Oct 06 '17 at 17:01
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    @Pooria My pleasure ... love this stuff ... – Bram28 Oct 06 '17 at 17:01