# Truth table of $\rightarrow$ given assumptions

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?

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.

• What might the condition be, do you have any ideas?
– Alex
May 14, 2023 at 23:58
• 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.
– zaq
May 15, 2023 at 14:29
• I like your suggestion, thanks!
– Alex
May 16, 2023 at 13:29