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I apologize if the title of this question can be better formulated, but I recently encountered a situation where what I thought was valid substitution led to an incorrect thing—at least I think.

I was trying to show myself why double negation elimination is not a theorem but how triple negation and single negation are equivalent. The former, $\neg\neg C \rightarrow C$, can be symbolized as

$((C \rightarrow \bot) \rightarrow \bot) \rightarrow C \tag{1}.$

which is just a special case of

$((C \rightarrow B) \rightarrow B) \rightarrow C \tag{2}.$

The latter, $\neg\neg\neg A \leftrightarrow \neg A$, can be symbolized as

$(((A \rightarrow \bot) \rightarrow \bot) \rightarrow \bot) \leftrightarrow (A \rightarrow \bot) \tag{3}.$

which is a special case of

$(((A \rightarrow B) \rightarrow B) \rightarrow B) \leftrightarrow (A \rightarrow B) \tag{4}.$

I thought it was valid to substitute $C$ for $(A \rightarrow B)$ in $(4)$ which would then reduce $(4)$ to

$((C \rightarrow B) \rightarrow B) \leftrightarrow C \tag{5}.$

Obviously assuming $(2)$ isn't a theorem means that $(5)$ can't be proven thereby contradicting the fact that triple negation is equivalent to single negation.

Does this mean that one can never substitute "across" connectives?

Edit

It's important to state that the reason I settled to prove the general forms (which is obviously much stronger than special cases) is due to an answer by Rob Arthan where it seems to me that he has a proof of $(4)$.

philomathic_life
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1 Answers1

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Just because $((C \rightarrow B) \rightarrow B) \leftrightarrow C$ can't be proven in general doesn't mean it can't be proven for particular values of $B$ and $C$.

The problem is that you have the implication backwards. If we have a proposition with a free variable $P(x)$ that's provable, then for any particular value $A$ (modulo variable capture), P(A) is also provable (by substitution). This implication doesn't go the other direction. That means when we negate, we don't get the implication "if $P(x)$ is not provable, then $P(A)$ is not provable". We do have the implication "if $P(A)$ is not provable, then $P(x)$ is not provable".

SCappella
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  • According to Rob Arthan, $(4)$ is provable which is equivalent to $(5)$, no? I understand that not being able to prove something for the general case doesn't mean it can't be proven for specific cases. I should admit that I don't entirely understand Rob's syntax which is why I specified Hilbert calculus. – philomathic_life Dec 12 '19 at 18:20
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    Why do you think $(4)$ is equivalent to $(5)$? The implication goes $(5) \to (4)$ but not vice versa. – SCappella Dec 12 '19 at 18:27
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    Because I'm extremely ignorant in formal logic especially intuitionistic logic, lol. Without getting technical—I know that violates the very nature of a formal logic—I am treating substitution as a form of "equality" and thus thought $(4) \rightarrow (5)$ as well. Thank you very much for highlighting the source of my problem. This is yet another example of my escaping the formal nature of logic by taking a "shortcut" by relying on my intuition if that makes sense. I'll work more on the basics. – philomathic_life Dec 12 '19 at 18:40
  • On a slight tangent, do you believe I should just delete this question? I don't want to "rob" you of reputation points, but I don't want to contribute to "garbage" either. I feel like my ignorance is preventing me from being able to accurately assess the quality of this question. – philomathic_life Dec 12 '19 at 18:46
  • @basketballfan22 if asking the question helped you understand something new, it's probably worth keeping. – SCappella Dec 13 '19 at 02:59