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I'm trying to prove the following tautologies:

\begin{align} & ⊢ (A \to (B \to A)) \\ & ⊢ ((A \to B) \to A) \to A \end{align}

For the first one, what I did was:

  1. $A$ assumption
    1. $B$ assumption
    2. $A$ restated
  2. $B \to A$ arrow-introduction of 2 and 3
    1. $A \to (B \to A)$ arrow-introduction of 1-4

But this doesn't look right to me.

For the second one, here's what I have so far:

  1. ((A --> B) --> A) assumption
    1. A --> B assumption
    2. A arrow-elimination of 1 and 2
Git Gud
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user146767
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4 Answers4

4

You'll be starting by assuming $(A\to B)\to A$.

Ideally one would want to prove $A\to B$ and then $A$ would follow, but of course $A\to B$ is not a tautology, so it can't be proven. But $\neg (A\to B)\lor (A\to B)$ is a tautology and therefore it can be proven.

So you'll prove the aforementioned tautology, then you'll perform $\lor\text{-Elim}$ on this tautology.

One of the cases follows from the initial hypothesis.

It remains to be shown that you can prove $A$ from $\neg(A\to B)$.

This goes something like this:

  • $\neg (A\to B)$
  • $\,\,\,\,\neg A$ (Hypothesis)
  • $\,\,\,\,\,\,\,\,\,\,\,\, A$ (Hypothesis)
  • $\,\,\,\,\,\,\,\,\,\,\,\,\bot$
  • $\,\,\,\,\,\,\,\, \,\,\,\,B$
  • $\,\,\,\,A\to B$
  • $\,\,\,\, \bot$
  • $\neg \neg A$
  • $A.$
Git Gud
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  • I'm not following the ¬(A→B) part. You show the contradiction, but then you go ahead and use A afterward... – user146767 Apr 30 '14 at 01:36
  • @user146767 Which line troubles you? – Git Gud Apr 30 '14 at 01:36
  • So I understand the not-A then A hypotheses (though I don't quite see the purpose of that yet), so of course it's a contradiction. But where does B come from, and how do you get A --> B after? – user146767 Apr 30 '14 at 01:38
  • Nice. I used contraposition, modus tollens, now you're explicitly appealing to excluded middle. I wonder if the formula we're both trying to prove is intuitionistically valid....If it's not then I wouldn't feel so bad for not being able to come up with a more 'direct' proof (if that makes sense). – Hunan Rostomyan Apr 30 '14 at 01:39
  • Do you have $\bot\text{-Elim}$? – Git Gud Apr 30 '14 at 01:39
  • @HunanRostomyan I highly doubt it is intuitionistically valid. – Git Gud Apr 30 '14 at 01:40
  • @user146767 Note that I edited my previous comment, it had a typo. – Git Gud Apr 30 '14 at 01:42
  • Which one? I didn't see a response – user146767 Apr 30 '14 at 01:44
  • @user146767 Click here, (no need to open on new tab). – Git Gud Apr 30 '14 at 01:46
  • oh, the question, I don't know what ⊥-Elim is – user146767 Apr 30 '14 at 01:50
  • @user146767 See the second proof in my answer. – Hunan Rostomyan Apr 30 '14 at 01:55
  • @GitGud I can send you the screenshot if you like and remove your proof from my answer. I thought this way would be faster. – Hunan Rostomyan Apr 30 '14 at 01:56
  • @HunanRostomyan Keep it. Thanks. – Git Gud Apr 30 '14 at 01:58
  • @user146767 I don't think I can save my proof without $\bot\text{-Elim}$. I'll try again tomorrow. I'm off to bed soon. – Git Gud Apr 30 '14 at 02:00
  • @HunanRostomyan I don't think I can use contrapositive either – user146767 Apr 30 '14 at 02:10
  • @GitGud I can use lemmas as long as I prove them separately as well. Is there a way to prove ⊥-Elim? – user146767 Apr 30 '14 at 02:10
  • @user146767 $\bot\text{-Elim}$ is a deduction rule and it is not a rule you can prove from the others, I think. – Git Gud Apr 30 '14 at 02:12
  • @HunanRostomyan Peirce's law (what you've tried to prove) is NOT intuitionistically valid, even though it has NO negations in it at all, and makes NO reference to the law of the excluded middle. " Examples include not only the law of excluded middle p ∨ ¬p, but also Peirce's law ((p → q) → p) → p, and even double negation elimination." http://en.wikipedia.org/wiki/Intuitionistic_logic It follows that the semantics of intuitionistic logic cannot be bivalent. – Doug Spoonwood Apr 30 '14 at 02:15
  • @DougSpoonwood Ah. That sounds right. I remember reading either Kolmogorov or Glivenko about the unacceptability of (double) negation elimination. I just didn't realize the connection to what you called 'Peirce's law', to our theorem here, until Git Gud produced a proof based on the excluded middle. Apparently they're all connected and intuitionistically unacceptable! Thank you very much for the comment. – Hunan Rostomyan Apr 30 '14 at 02:20
  • @HunanRostomyan They aren't all connected. If you look at the proof of Peirce's law in say a conditional-negation axiomization of classical logic (or even the similarly structure natural deduction system partially outlined in my answer), and forbid inter-definability of connectives as intuitionists do, then a proof of Perice's law says nothing about the law of the excluded middle, which is a disjunction. On the other hand, if you allow for interdefiniability of connectives, since Cpq:=ANpq, from the weak law of identity Cpp we can obtain ANpp and then ApNp the law of the excluded middle. – Doug Spoonwood Apr 30 '14 at 02:35
3

You've solved the first, so here are two ways of proving the second one. 'Taut Con' is modus tollens.

enter image description here

I've taken the liberty of typing Git Gud's alternative proof more explicitly here (please vote his answer up if you too find it insightful). It highlights the connection to the law of excluded middle and is economical in its use of special rules (e.g. doesn't appeal to modus tollens). Here it is:

enter image description here

Another way would be to prove the contrapositive, using just $\rightarrow$-intro & Reit:

enter image description here

Community
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Hunan Rostomyan
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1

I use Polish notation. The second formula becomes CCCabaa. The rule I'll use (other than arrow introduction, which I'll call "Ci", and detachment/conditional elimination, which I'll call "Co") is {CN$\alpha$$\beta$, CN$\alpha$N$\beta$} $\vdash$ $\alpha$, which I'll call "No".

hypothesis 1  |     CCaba
hypothesis 2  ||    Na
hypothesis 3  |||   a
hypothesis 4  ||||  Nb
hypothesis 5  ||||| a
Ci 5-5     6  ||||  Caa
Co 6, 3    7  ||||  a
Ci 4-7     8  |||   CNba
hypothesis 9  ||||  Nb
hypothesis 10 ||||| Na
Ci 10-10   11 ||||  CNaNa
Co 11, 2   12 ||||  Na
Ci 9-12    13 |||   CNbNa
No 13, 8   14 |||   b
Ci 3-14    15 ||    Cab
Co 1, 15   16 ||    a
Ci 2-16    17 |     CNaa
hypothesis 18 ||    Na
Ci 18-18   19 |     CNaNa
No 17, 19  20 |     a
Ci 1-20    21       CCCabaa

For the first one you could also write:

hypothesis 1 |   a
hypothesis 2 ||  b
hypothesis 3 ||| a
Ci 3-3     4 ||  Caa
Co 4, 1    5 ||  a
Ci 2-5     6 |   Cba
Ci 1-6     7     CaCba

I have to wonder why you didn't also get asked to prove CCpCqrCCpqCpr. Since you can prove CCpCqrCCpqCpr using only conditional introduction and detachment, a consequence here come as that all implications (well-formed formulas with just "C"'s and variables in them) which qualify as tautologies can get proved from conditional introduction, detachment, and the rule

{CN$\alpha$$\beta$, CN$\alpha$N$\beta$} $\vdash$ $\alpha$, since {CpCqp, CCpCqrCCpqCpr, CCCpqpp} under detachment and uniform substitution comes as sufficient for the implicational propositional calculus.

Doug Spoonwood
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0

Pierce's Law: $$\def\fitch#1#2{~~~~\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{}{\fitch{~~1.~(A\to B)\to A}{\fitch{~~2.~\neg A}{\fitch{~~3.~A}{~~4.~\bot\hspace{13.5ex}{\neg}\mathsf E~3,2\\~~5.~B\hspace{14ex}\mathsf X~4}\\~~6.~A\to B\hspace{11.5ex}{\to}\mathsf I~3{-}5\\~~7.~A\hspace{17ex}{\to}\mathsf E~6,1\\~~8.~\bot\hspace{17ex}\neg\mathsf E~7,2}\\~~9.~\neg\neg A\hspace{17ex}\neg\mathsf I~2{-}8\\10.~A\hspace{20ex}{\neg\neg}\mathsf E~9}\\11.~((A\to B)\to A)\to A\hspace{3ex}{\to}\mathsf I~1{-}10}$$

Graham Kemp
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