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I am reading the book by Kleene "Introduction to Metamathematics". In Chapter 2 Paragraph 7 on "Mathematical Induction" there is a problem to prove the following lemma.

Lemma 1: A proper pairing of $2n$ parentheses ($n>0$ and $n$ is a natural number) contains an innermost pair, i.e. a pair which includes no other of the parentheses between them.

Following definitions are used:

Proper pairing - one-to-one pairing between $n$ left parentheses "$($" and $n$ right parentheses "$)$" such that for each pair left parenthesis is on the left from the right parenthesis and if no two pairs separate each other.

Pairs of parentheses that separate each other - two pairs separate each other if they occur in the order $(_i (_j )_i )_j$ (ignoring everything else).

To prove this lemma Kleene suggests to use the principle of course-of-values induction, which I assume is known to you.

I wanted to try to prove this lemma using "ordinary" induction. What I would like to ask is whether my proof is correct, and, if it is not, where is the mistake? Also, any comments are appreciated.

  1. Consider the basis step, i.e. when $n=1$. Then, there are only two possible arrangements of parenthesis, namely, $()$ and $)($. The latter is not proper, therefore, there is only one proper pairing. This pair satisfies the definition of an innermost pair, so the lemma is true in this case.

  2. Assume that if $n = m$ then the lemma holds. Consider $n = m+1$, which means that there is given a proper pairing of $2m+2$ parentheses. Consider one (arbitrary) pair. Either it is innermost or it is not (here, I am using law of excluded middle, but since this theorem is supposed to be in metamathematics where there is no axioms a priori, and, thus, I feel that theorems in metamathematics should be just some statements with some justifiable reason to be assumed, this is okay. In other words, there is no law of excluded middle in metamathematics and the argument is considered valid due to the construction of parentheses and our intuition). If it is innermost then the lemma is true. If it is not, then consider remaining $2m$ parentheses. By induction principle, we have that pairing between these parentheses has an innermost pair. Now, it could be that for these $2m$ parentheses it is an innermost pair, but adding the additional pair, it is not innermost anymore. If it still is the innermost pair, then the lemma is true. Otherwise, it should be the case that at least one of two additional parentheses is in this pair. It couldn't be one of the parentheses because then the pairing would not be proper. So, then, there must be both of the parentheses inside this pair. That would imply that the additional pair is innermost pair. But that is a contradiction as we assumed the pair that is removed and added is not innermost. It must be then that the innermost pair of remaining $2m$ parentheses is indeed innermost also for $2m+2$ parentheses. In any case, there is an innermost pair.

Therefore, by the principle of induction, the lemma is true for all natural numbers.

Daniels Krimans
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    I don't see anything wrong with what you wrote. Am I right that it's more obvious there is an outermost pair? I think my proof strategy would be to eliminate the outermost pair, and then use the induction hypothesis. I feel like neither your proof nor my "proof" uses strong induction (which Wikipedia tells me is the same as "course of values induction"), both our proofs seem to just use ordinary induction. So maybe there is some subtlety I'm missing. – CJD Jul 10 '18 at 20:48
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    @CJD I don't think it is obvious that there is the outermost pair. Consider proper pairing " () () ". Then, there is no such outermost pair (if I correctly interpreted what you meant). – Daniels Krimans Jul 10 '18 at 21:10
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    Thanks for that example. I agree it looks like there is no outermost pair, but to me it also looks like there is no innermost pair. I think you could define outermost pair by "not between any other parentheses". How about this variant of the proof you gave: Look at the parentheses inside the pair you chose. That would be a way to use strong induction. – CJD Jul 10 '18 at 21:26
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    It seems easier to just consider the rightmost left parenthesis, say $(_A$. The right parenthesis paired with it $)_A$ is to its right and there can't be another left parenthesis in between because $(_A$ was rightmost. Could another right parenthesis $)_B$ be in between? Its paired left parenthesis $(_B$ would be left of $(_A$, again because $(_A$ is rightmost, So we'd have $(_B(_A)_B)_A$, an illegal separation. So $(_A$ and $)_A$ are adjacent. – Andreas Blass Jul 11 '18 at 03:14
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    If you really, really want to use induction, you might use it to prove that there is a rightmost left parenthesis, and then proceed as in my previous comment. – Andreas Blass Jul 11 '18 at 03:16

1 Answers1

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Your proof is ok: its critical step is the observation that if the innermost pair of parenthesis doesn't remain innermost after adding the removed pair, then the removed pair must have been innermost.

Using strong induction, just take a proper pairing $a_1\dots a_{2n}, \ a_i\in \{{\bf(} , \, {\bf)}\}$ and assume the hypothesis for all $n<m$.
Now if, say, $a_k$ is the pair of $a_1$, then $a_2\dots a_{k-1}$ inherits a proper pairing, thus - applying the hypothesis for $n=\frac{k-2}2\, $ - gives an innermost pair.

Berci
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