3

This is all in the context of a Hilbert system with modus ponens $A, A\to B \vdash B$ and axioms:

Axiom $1$: $A \to (B \to A)$

Axiom $2$: $(A \to (B \to C)) \to ((A \to B) \to (A \to C))$

Axiom $3$: $(\lnot B \to \lnot A) \to (A \to B)$

So I tried to prove double-negation elimination by doing the following:

\begin{array}{ccc} 1 & \{ \lnot \lnot a \} \vdash \lnot \lnot a & \text{Premise}\\ 2 & \{ \lnot \lnot a \} \vdash \lnot \lnot a \to (\lnot \lnot \lnot \lnot a \to \lnot \lnot a) & \text{ Axiom } 1\\ 3 & \{ \lnot \lnot a \} \vdash \lnot \lnot \lnot \lnot a \to \lnot \lnot a & \text{ Modus Ponens } 1, 2\\ 4 & \{ \lnot \lnot a \} \vdash (\lnot \lnot \lnot \lnot a \to \lnot \lnot a) \to (\lnot a \to \lnot \lnot \lnot a) & \text{ Axiom } 3\\ 5 & \{ \lnot \lnot a \} \vdash \lnot a \to \lnot \lnot \lnot a & \text{ Modus Ponens } 3, 4\\ 6 & \{ \lnot \lnot a \} \vdash (\lnot a \to \lnot \lnot \lnot a) \to (\lnot \lnot a \to a) & \text{ Axiom } 3\\ 7 & \{ \lnot \lnot a \} \vdash \lnot \lnot a \to a & \text{ Modus Ponens } 5, 6\\ 8 & \{ \lnot \lnot a \} \vdash a & \text{ Modus Ponens } 1, 7\\ 9 & \vdash \lnot \lnot a \to a & \text{ Deduction Theorem } \\ \end{array}

And then I continued on, to prove double negation introduction:

\begin{array}{ccc} 10 & \vdash \lnot \lnot \lnot a \to \lnot a & \text{Double Negation Elimination}\\ 11 & \vdash (\lnot\lnot \lnot a \to \lnot a) \to (a \to \lnot \lnot a) & \text{ Axiom } 3\\ 12 & \vdash a \to \lnot \lnot a & \text{ Modus Ponens } 10, 11\\ \end{array}

My questions:

  1. Am I allowed to invoke the deduction theorem in the middle of a proof like this (step $9$)? Is this considered a context change? Is context-changing allowed?

  2. Are the proofs even right/correct?

  3. Assuming it's all correct: Are we able to combine them to say that $a \to \lnot \lnot a$ and $\lnot \lnot a \to a$, therefore $a \iff \lnot \lnot a$ or is this symbol not allowed?

  4. What ultimately let us say $\lnot \lnot a = a$, the use of the equal sign here?

Taroccoesbrocco
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user525966
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1 Answers1

2

  1. The sequence 1-8 is a derivation (in the Hilbert system for implicative propositional classical logic) of $\lnot \lnot a \vdash a$. Step 9 does not correspond to an inference rule in such Hilbert system, thus the sequence 1-9 is not a derivation of $\vdash \lnot \lnot a \to a$ in such Hilbert system. The deduction theorem says that, since there is a derivation of $\lnot \lnot a \vdash a$, then there exists another derivation (which you didn't construct, but there exists) of $\vdash \lnot \lnot a \to a$. So, if the question is "Show a derivation of $\vdash \lnot \lnot a \to a$ in such Hilbert system", you didn't answer it; but if the question is "Show that $\vdash \lnot \lnot a \to a$ is derivable in such Hilbert system", your answer is fine (except that step 9 is not allowed as an inference rule in the Hilbert system, i.e. you should consider the sequence 1-8 and then apply the deduction theorem to conclude). See also here.

  2. Concerning the proof of $\vdash \lnot \lnot a \to a$, see point 1. Your derivation for $\vdash a \to \lnot \lnot a$ (which is completely independent and separated from the derivation 1-8) is correct, on the proviso that you have already proved that $\vdash \lnot \lnot \lnot a \to \lnot a$ is derivable in the Hilbert system.

  3. If you want to consider $\leftrightarrow$ as a connective in the formal language, then you need to add some axioms concerning it in the Hilbert system. Without these axioms, the formula $a \leftrightarrow \lnot \lnot a$ makes no sense in the Hilbert system and hence it is not allowed. The list of axioms that you have to add to the Hilbert system for the connective $\leftrightarrow$ is here.
  4. The notation $a = \lnot \lnot a$ (I prefer to write $a\equiv \lnot \lnot a$ or $a\dashv\vdash \lnot \lnot a$) means that both $\vdash \lnot \lnot a \to a$ and $\vdash a \to \lnot \lnot a$ are derivable in the Hilbert system, which is essentially what you have proved (or want to prove). Note that $a = \lnot \lnot a$ and $a\equiv \lnot \lnot a$ and $a\dashv\vdash \lnot \lnot a$ are not formulas in the language, they are just notations in the meta-language. See also point 3.
Taroccoesbrocco
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  • So would step $9$ technically need to be a separate proof, a one-liner invoking deduction theorem from the last line ($8$) of the first proof? – user525966 Sep 23 '18 at 18:43
  • Yes, and it is not a derivation but a proof (in the meta-language) that such a derivation exists. – Taroccoesbrocco Sep 23 '18 at 18:45
  • If I literally wanted to show the derivation/proof explicitly then I would need to start off with $\emptyset \vdash$ without using the deduction theorem? Is it even considered a valid "proof" then to use the theorem and simply say "therefore this proof exists" without explicitly showing it? I imagine most would be "convinced" but is it "enough" for a proof in practice? – user525966 Sep 23 '18 at 18:49
  • AFAIK in computer science, boolean algebra, things like that, to my knowledge it is allowable to write $a = \lnot \lnot a$ so is this more of a... using the result from formal logic in some practical sense in another situation/context using helpful notation, on a meta level? – user525966 Sep 23 '18 at 18:52
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    @user525966 - It depends on the question and on your professor. Anyway, consider that in general questions ask to prove derivability, not a concrete derivation. Without the deduction theorem, the Hilbert system is awkward and innatural. Moreover, the proof of the deduction theorem is constructive, so if you apply the proof of the deduction theorem to your derivation 1-8, you get an explicit derivation of $\vdash \lnot \lnot a \to a$ in the Hilbert system. – Taroccoesbrocco Sep 23 '18 at 18:52
  • @user525966 - Yes, it is. – Taroccoesbrocco Sep 23 '18 at 18:53
  • (I am not a student and have no professor -- when I ask "is it enough" I mean in a sort of "practical" sense? Like if you asked me to prove that $\lnot \lnot a \to a$ and I gave you those steps and used the deduction theorem, would people nod at this and say yes, I'm convinced now, or would they go oh no no wait a second, you didn't technically prove it yet by showing all the steps because you only showed the proof exists, not the actual proof itself) – user525966 Sep 23 '18 at 18:56
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    @user525966 - In this case, if you show me the derivation 1-8 and then claim that $\vdash \lnot \lnot a \to a$ is derivable thanks to the deduction theorem, I'm perfectly satisfied. – Taroccoesbrocco Sep 23 '18 at 19:01
  • @user525966: You are correct that if you want an actual proof in the Hilbert-style system then you cannot use the deduction theorem. To make things clear, what you have written is not a proof in the Hilbert-system but rather an explanation in the meta-system that there exists a proof in the Hilbert-system. Whether people are convinced that there really is a proof depends on whether they are convinced that the deduction theorem is in fact true. Proving the deduction theorem requires induction, so one who does not believe induction will likely not be convinced. – user21820 Sep 24 '18 at 05:06
  • @user525966: Such a fine distinction between writing down an actual proof and giving an explanation of existence of a proof may seem weird, but it is actually crucial in logic, especially first-order logic. However strange, it may be that you can prove the existence of a proof but cannot actually write the proof down. Worse still, it may even be that you can prove the existence of a proof but actually none exists! But such interesting phenomena will have to wait until you learn first-order logic. =) – user21820 Sep 24 '18 at 05:11