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Say we have a (positively) real valued logic, i.e. formulas evaluate to the extended interval $[0,\, \infty]$. Furthermore, we have a Gödel implication $a \rightarrow b$ that evaluates to $\infty$ wherever $a$ evaluates to something less-than or equal $b$, and to $b$ otherwise, i.e. the semantics $[[a \rightarrow b]]$ of the Gödel implication under a model $m$ is given by $$[[ a \rightarrow b]](m) ~{}={}~ \begin{cases}\infty~, & \text{if } [[ a ]](m) \leq [[b]](m)\\ [[b]](m)~, & \text{otherwise.}\end{cases}$$

In classical logic, we can think of $a \rightarrow \:{\cdot}\:$ as a kind of "assume $a$" operation. How can we think of $5 \rightarrow x^2$ as a kind of "assume 5"? How is $5 \rightarrow x^2$ an assumption of $5$ on $x^2$?

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    $\begingroup$ "In classical logic, we can think of $a\rightarrow\bullet$ as a kind of 'assume $a$' operation." I'm not sure exactly what you mean by this, but any way I can think of to make this statement precise will equally well apply to the Gödel implication. $\endgroup$ Commented Aug 12, 2023 at 13:25
  • $\begingroup$ Can you elaborate on ways you can think about making this statement precise and how they apply to Gödel implication? $\endgroup$
    – blk
    Commented Aug 12, 2023 at 16:37

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Given formulas $\varphi$ and $\psi$, what is the role of the formula $\varphi\to \psi$?

The obvious thing is modus ponens: if we know $\varphi\to \psi$ and we know $\varphi$, then we can conclude $\psi$.

But this property does not completely characterize $\varphi\to\psi$: other formulas also play this role. For example, if we know $\psi$ and we know $\varphi$, then we can conclude $\psi$. Or if we know $\bot$ and we know $\varphi$, then we can conclude $\psi$.

Slightly less obviously, $\varphi\to \psi$ is the weakest formula with the modus ponens property. That is, if $\chi$ is any formula such that if we know $\chi$ and we know $\varphi$, then we can conclude $\psi$, then $\chi$ entails $\varphi\to \psi$. To put it another way, if in our current state of knowledge ($\chi$), additionally knowing $\varphi$ would allow us to conclude $\psi$, then our current state of knowledge entails $\varphi\to \psi$.

Now let's write $[\varphi]$ for the truth value of $\varphi$, so that $[\varphi]\leq [\psi]$ whenever $\varphi$ entails $\psi$, and $[\varphi\land \psi] = [\varphi]\wedge [\psi]$, where the first $\wedge$ is logical "and" and the second wedge is the meet (greatest lower bound) in the lattice of truth values. How should $[\varphi\to\psi]$ depend on $[\varphi]$ and $[\psi]$?

Translating the above discussion to truth values, we have that $[\varphi]\land [\varphi\to \psi]\leq [\psi]$, and whenever $[\varphi]\land [\chi]\leq [\psi]$, then $[\chi]\leq [\varphi\to\psi]$. That is, $[\varphi\to\psi]$ is the greatest truth value $p$ with the property that $[\varphi]\wedge p \leq [\psi]$. This universal property is exactly how $\to$ is defined in Heyting algebras.

In your case, where the truth values are the linearly ordered set $[0,\infty]$, we find that if $a\leq b$, then every $c$ has the property that $a\wedge c \leq b$. Thus the greatest such $c$ is the maximum element $\infty$. And if $a > b$, then $a\wedge c \leq b$ if and only if $c\leq b$. So the greatest such $c$ is $b$ itself. This explains the definition of the Gödel implication.

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  • $\begingroup$ Thanks Alex, this is a nice account of the semantics of Gödel implication and its adjoint $\wedge$. But I don't understand how this answers my question. I am aware of all of the above and still have the question: In which way can I think of $\varphi \rightarrow {\cdot}$ as "making the assumption $\varphi$"? It might be obvious, but I don't see it. $\endgroup$
    – blk
    Commented Aug 13, 2023 at 23:53
  • $\begingroup$ @blk Well, can you explain how precisely you think of $\varphi\to \bullet$ as "making the assumption $\varphi$ in classical logic? Then maybe I would have some sense of the kind of explanation you're looking for. $\endgroup$ Commented Aug 14, 2023 at 0:12
  • $\begingroup$ This may be a bit naive, but say I want to prove $\psi$. Then I have to consider only those cases where $\varphi$ holds. For cases where $\varphi$ does not hold, I don't have to prove anything (because I assume that $\varphi$ is definitely true). $\endgroup$
    – blk
    Commented Aug 14, 2023 at 0:15
  • $\begingroup$ @blk I think what you're expressing in your last comment is that proving $\varphi\to \psi$ is the same as proving $\psi$, but considering those cases where $\varphi$ holds. This is just a restatement of the adjointness: $\chi$ entails $\varphi\to \psi$ if and only if $\chi\land \varphi$ entails $\psi$. $\endgroup$ Commented Aug 14, 2023 at 14:41

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