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For a while I've been wondering if it's possible arrive at the truth table for $\rightarrow$ from the following definition for $\leftrightarrow$

$p\leftrightarrow q :\Leftrightarrow (p\rightarrow q) \wedge (q \rightarrow p)$

and these assumptions

  1. $p\leftrightarrow q$ is true whenever both $p$ and $q$ have the same truth value.
  2. The truth table for conjunction $\wedge$ is the standard one.
  3. Out of all the 16 possible binary connectives, $\rightarrow$ is neither $\leftrightarrow$, $\wedge$ nor $\vee$.

So far I have not been able to arrive at the answer. However, Assumption 3 does give the hint that we have 13 truth tables to go through. Do we need other "obvious" natural assumptions to arrive at the actual truth table for $\rightarrow$? If so, what are they?

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No, those conditions aren't sufficient to uniquely determine $\rightarrow$. For instance, replacing the meaning of $p \rightarrow q$ with $q \rightarrow p$ gives a second possibility.

Now, consider that in the expression $p \leftrightarrow q := (p \rightarrow q) \land (q \rightarrow p)$ each of the $\land$-ed factors "provides" exactly one row of the truth table that is equal to $0$. This is because we must have a total of two such rows (for $\neg p, q$ and $p, \neg q$) and neither term can provide both: otherwise, $\rightarrow$ would be equivalent to $\leftrightarrow$, violating condition $3$. Other rows must be $1$ in both terms, as otherwise we would have a $0$ in one of those rows after $\land$-ing.

This actually shows that there are exactly two solutions, as we can really only permute which term makes which row $0$ and there are $2$ rows for $2$ terms, so simply $2! = 2$ possibilities.

BTW, it seems that most of condition $3$ is redundant. We can easily check that $\land$ and $\lor$ don't satisfy the initial definition of $\leftrightarrow$, so we only really need that $\rightarrow$ isn't $\leftrightarrow$.

So, to uniquely define $\rightarrow$, you need one more condition to eliminate of the two possibilities.

It's actually fairly difficult to find a simple, intuitive condition which does this, but doesn't characterize $\rightarrow$ specifically (I guess this is what you mean when you say you want avoid the "obvious" conditions). This is because we're basically trying to differentiate between $\rightarrow$ and $\leftarrow$.

I guess the most straightforward condition would be "if $p$ is false and $q$ is true, $p \rightarrow q$ is true". But that's not very intuitive.

A pretty property of $\rightarrow$ that would suffice is modus ponens: $(p \land (p \rightarrow q)) \rightarrow q$ is a tautology. I think this is maybe the prettiest option and you can check that this is not true if $\rightarrow$ is replaced with $\leftarrow$. Still, modus ponens is a bit stronger than necessary, as it guarantees that $0 \rightarrow 0$ is true, but that's redundant to us because it follows from the definition of $\leftrightarrow$.

Still, if you want a condition that doesn't involve other connectors, you can try requiring that $p \rightarrow (q \rightarrow p)$ is a tautology. This is another important property of $\rightarrow$, but it is also a bit redundant because of its strength, unfortunately.

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  • $\begingroup$ What might the condition be, do you have any ideas? $\endgroup$
    – Alex
    May 14, 2023 at 23:58
  • $\begingroup$ I've edited my post to comment on this. I'd probably go with modus ponens, but it's obviously a matter of preference. Also, you don't need condition $3$ at all if you add this new condition to differentiate between $\rightarrow$ and $\leftarrow$, as this will then certainly eliminate $\leftrightarrow$ as a possibility. $\endgroup$
    – zaq
    May 15, 2023 at 14:29
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    $\begingroup$ I like your suggestion, thanks! $\endgroup$
    – Alex
    May 16, 2023 at 13:29

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