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I'm reading Dalen's Logic and Structure.

He gives the definition of the set $PROP$:

Definition $\bf2.1.2$ The set $PROP$ of propositions is the smallest set $X$ with the properties

$\begin{array}{rl} \rm(i)&p_i\in X(i\in N),\bot\in X,\\ \rm(ii)&\varphi,\psi\in X\Rightarrow(\varphi\wedge\psi),(\varphi\vee\psi),(\varphi\to\psi),(\varphi\leftrightarrow\psi)\in X,\\ \rm(iii)&\varphi\in X\Rightarrow(\neg\varphi)\in X.\\ \end{array}$

Then there are some examples:

enter image description here

And he says that $¬¬\bot \not\in PROP$, but I don't understand it. Isn't $¬¬ \bot=\bot$ due to the property of double negation? Considering this is true, $\bot\in PROP$. Of course, in this question, I'm pressuposing some kind of evaluation, that $¬¬ \bot$ becomes $\bot$ due to the law of the double negation, but perhaps it should be hold in an unevaluated form.

Ex:. $1+1=2$ because we presupose the $1$ merges with $1$ and then becomes two. If the merge (evaluation) does not happen, then I guess:

$$1+1=1+1$$ $$2=2$$ $$2\neq 1+1$$ $$1+1\neq 2$$

Perhaps it's something like that, I took the idea from Mathematica. Sometimes it does treats things as different in the absence of full evaluation.

Hakim
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Red Banana
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3 Answers3

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As the author himself alerts to: look at the brackets. What's in $PROP$ is $\left(\neg \left(\neg\bot\right)\right)$ (note the two pairs of $()$).

It isn't true that $\neg \left(\neg \bot\right)=\neg \neg \bot$ simply because they aren't the same string of characters. In mathematical logic, particularly in formal systems, one looks at propositions as actual mathematical objects, more precisely as finite strings of characters. The second character in $\neg \left(\neg \bot\right)$ is $\color{blue}($ which differs from the second character in $\neg \neg \bot$. Therefore they are different.

It is likely that eventually, after some familiarity with these concepts has been acquired,that $\neg \neg \bot$ might be defined as $\left(\neg (\neg \bot)\right)$, but for now they are different. Not only that, as the author proves, $\neg \neg\bot \not \in PROP$ according to the formation rules.

Regarding your example, if you look at $1+1$ and $2$, not as representing objects, but as objects themselves, then they are different because, well..., they are different.

Git Gud
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  • But isn't $\neg \left(\neg\bot\right)=¬¬ \bot = \bot$? See what I added. – Red Banana Apr 10 '14 at 21:46
  • @IgäriaMnagarka Since you asked what's in the comment, in the question, I replied in my answer. – Git Gud Apr 10 '14 at 21:51
  • Got it. So the background for it is similar to the one I gave in the question. – Red Banana Apr 10 '14 at 21:53
  • Yes. If they're seen as strings without rules of interaction, I guess. – Red Banana Apr 10 '14 at 21:54
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    @IgäriaMnagarka Yes, you shouldn't interpret them.They are just symbols. They could have very well used $\huge ☺$ instead of $\neg$ and then you wouldn't have had this problem. – Git Gud Apr 10 '14 at 21:56
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    As far as I can see, $\neg(\neg\bot)$ isn't in $PROP$ either; what is in $PROP$ is $(\neg(\neg\bot))$. Each $\neg$ needs a pair of parentheses. – Andreas Blass Apr 10 '14 at 22:18
  • @AndreasBlass Yes, of course. Thank you. – Git Gud Apr 10 '14 at 22:19
  • @IgäriaMnagarka Please read Andreas' comment above. I fixed my answer. – Git Gud Apr 10 '14 at 22:20
  • @AndreasBlass Why each $¬$ needs a pair or parentheses? – Red Banana Apr 10 '14 at 23:15
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    @IgäriaMnagarka $(i)$ gives you $\bot \in X$, then in $(iii)$ replacing $\varphi$ with $\bot$ gives you $\bot \in X\implies (\neg\bot )\in X$, therefore $(\neg\bot)\in X$. Now using $(iii)$ again, this time with $\varphi=(\neg \bot)$ yields $(\neg \bot)\implies (\neg (\neg \bot))\in X$, therefore $(\neg (\neg \bot))\in X$. – Git Gud Apr 10 '14 at 23:19
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    @IgäriaMnagarka More generally than Git Gud's last comment: Look at the definition of $PROP$ in the question. The only clause that lets you introduce a $\neg$, namely clause (iii), also requires you to introduce a pair of parentheses. – Andreas Blass Apr 10 '14 at 23:24
  • Yes. Thanks guys! – Red Banana Apr 10 '14 at 23:26
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    This all said if the author had defined PROP with iii as "if $\phi$ belongs to PROP, then ¬$\phi$ belongs to PROP" no ambiguities similar to how ambiguities would occur if you dropped the parentheses in the second clause. Also, in that case ¬¬⊥ would belong to PROP, but (¬(¬(⊥)) would not belong to PROP. – Doug Spoonwood Apr 11 '14 at 04:07
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Regarding your example :

$1+1=2$

if you want to "test it" according to van Dalen's rules, you must first take into account that it is a formula of predicate logic and, in particular, of formal number theory.

To show that it is well-formed, you must review the formation rules of the language; see Dirk van Dalen, Logic and Structure (5th ed - 2013), page 56-57 :

Definition 3.3.1 TERM is the smallest set $X$ with the properties [...]

<p><strong>Definition 3.3.2</strong> <em>FORM</em> is the smallest set $X$ with the properties [...]</p>

Then, see page 82 : The language of arithmetic, and the notations introduced in page 83.

Mauro ALLEGRANZA
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I guess that you are reading the book I know from Dirk van Dalen. If that's the case, at some point the author states that:

In order to simplify our notation we will economize on brackets. We will always discard the outermost brackets and we will discard brackets in the case of negations. Furthermore we will use the convention that $\land$ and $\lor$ bind more strongly than $\rightarrow$ and $\leftrightarrow$ (cf. $\cdot$ and + in arithmetic), and that $\neg$ binds more strongly than the other connectives.

Page 12 of Dirk van Dalen's Logic and Structure - Fifth edition.

Maybe you skipped it and that's what generated the confusion.

nnov
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