The Renewal Equation

Responses

1. Tim McL

   April 3, 2024 at 7:17 pm

   How do you feel about a theorem like

   Th: If $x=2+2$, then either $x=4$ or $x=666$.

   Pf: Well, if $x=2+2$, then $x=4$, so either $x=4$ or $x=666$.

   We don’t need $x=666$ to be possible.

   This isn’t exactly your method, but is it connected to your objection?

   Just a thought.

   I certainly agree that there are plenty of places where ordinary language and formal logic diverge. A very simple one from Graham Priest’s Logic: A Very Short Introduction is this: We all agree that “$p$ and $q$” is equivalent to “$q$ and $p$.” Is “John hit his head and fell down” the same as “John fell down and hit his head?”

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   1. Olly

      April 3, 2024 at 8:52 pm

      Yes, I did notice that, as long as $p\implies r$, $q$ could be any statement, even a totally unrelated one. And I don’t really have an objection. I think this method is logically sound; it just feels like creating something out of nothing.

      Reply
   2. Olly

      April 3, 2024 at 9:29 pm

      Now I am wondering how you would express, in general terms, the idea that $p\implies (q\lor r)$ and both $q$ and $r$ are possibilities. I guess $[(p\land \lnot q)\implies r]\land [(p\land \lnot r)\implies q]$ might be one way. I think proving that would have been my instinct given the proposition about $x^2=2x$.

      Reply
      1. Tim McL

         April 6, 2024 at 4:55 pm

         The trouble with your translation is that $(p\wedge q)\rightarrow r$ doesn’t imply that $p\wedge q$ can be true:
         $$
         ((1+1=3)\land(1+2=26))\rightarrow(2+3=29)
         $$
         is true because you can add both equations.

         Worse, in classical propositional logic, $p$ and $q$ and $r$ are either true or false. There’s not a way to say “Sometimes true and sometimes false.”

         There are a few ways around that. One is to make $p$ and $q$ and $r$ into predicates that depend on some variable $x$. You could then write things like
         $$
         \exists x(p(x)\land\neg q(x))
         $$
         to mean that for some values of $x$, $p(x)$ is true and $q(x)$ is false. That’s probably what you want to do.

         A less mathematical approach might be to use modal logic, which I love. Here, in addition to True and False, you have the modes $p$ is Necessary ($\square p$) and $p$ is Possible ($\Diamond p$). With this notation, you could say things like
         $$
         \Diamond(p\land\neg q)
         $$
         to mean that it’s possible that $p$ is true and $q$ is false. The way logicians would usually interpret $\Diamond p$ this is to say that there’s a possible world in which $p$ is true, rather than that in some given world there are parameters that can make $p$ true.

         I find modal logic fascinating. You end up asking questions like, is $\Diamond\Diamond p\rightarrow\Diamond p$ a tautology? That is, if it’s possible that $p$ is possible, is $p$ in fact possible? Or is $\square p\rightarrow\square\square p$ a tautology? If $p$ is necessary, is it necessarily necessary? Or could $p$ just happen to be necessary without necessarily being necessary? Or what about $\Diamond p\rightarrow\square\Diamond p$? If $p$ is possible, does it have to be possible?

         There are systems of modal logic where all these things are true, and systems of modal logic where they’re false. It’s confusing logic (working inside the formal system) and metalogic (reasoning about the formal system from outside) to say this, but
         $$
         \Diamond(\Diamond\Diamond p\rightarrow\Diamond p)\land
         \neg\square(\Diamond\Diamond p\rightarrow\Diamond p).
         $$
         🤣

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