Can Curry's Paradox be stated in propositional logic as:
$[A\iff [A\implies B]] \implies B$
This is a tautology.
Note that:
$[A\iff [A\implies B]] \equiv A\land B$
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Dan Christensen
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If it is a tautology, it's kind of unlikely to be a paradox ..... – Peter Smith Jun 29 '18 at 19:25
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1@PeterSmith Some call $\neg A \implies [A \implies B]]$ a "paradox." It, too, is a tautology. – Dan Christensen Jun 29 '18 at 19:43
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@DanChristensen Who are those some? – Noah Schweber Jun 29 '18 at 20:46
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1@NoahSchweber See https://en.wikipedia.org/wiki/Paradoxes_of_material_implication – Dan Christensen Jun 29 '18 at 21:17
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@DanChristensen Ah. Regardless, it's a paradox on a very different level than Curry's paradox; we can neutralize Curry by rejecting material implication, but Curry's paradox is much more than just standard $\implies$-manipulation. – Noah Schweber Jun 29 '18 at 21:21
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No, this is not Curry's paradox. (I'm sure you're familiar with what I write below; I'm including it for completeness and the benefit of other readers.)
Curry's paradox is used to "prove" an arbitrary claim, assuming the "admissibility" of a certain kind of self-referential statement. Namely, given an arbitrary sentence $\varphi$, Curry examines the sentence $$(*)_\varphi\quad\mbox{If this statement is true, then $\varphi$}.$$ This has the form "$A\implies\varphi$," and in order for it to be false $A$ must be true. That much doesn't rely on self-reference.
However, where self-reference comes in is in the observation that this situation can't occur: $$\mbox{$(*)_\varphi$ is false $\implies$ the hypothesis of $(*)_\varphi$ is true $\iff$ $(*)_\varphi$ is true;}$$ so we get "$\neg(*)_\varphi\implies(*)_\varphi$," that is, $(*)_\varphi$ is true.
Now we simply apply modus ponens to get $\varphi$.
When cast as a theorem, Curry's paradox amounts to a limitation, in any nontrivial formal system, on either the kinds of self-reference expressible in that system or the extent to which the standard rules of logic hold for such self-referential sentences within that system.
As you observed, the sentence you write down is not paradoxical; it is missing the self-referentiality which is crucial to Curry's paradox. What it is is the observation that makes the Curry paradox "tick:" we apply it to the assumption that "This sentence implies $\varphi$" is expressible in our system. But it's wrong to say that this is Curry's paradox; rather, Curry's paradox relies on the validity of the principle. This is a worthwhile observation since in weak formal systems (e.g. I believe relevant logics) Curry's paradox can break down.
Noah Schweber
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2@DanChristensen Since classical logic is consistent, it avoids every paradox. – Noah Schweber Jun 30 '18 at 17:21
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You can try QSAT, in classical logic you get:
$$\vdash_C \forall P.(Q \leftrightarrow (Q \rightarrow P)) \rightarrow \forall P.P$$
One of the interesting challenges could be to show the same in non-classical logic. It then depends how you read the propositional quantifier. If you only assume this distributivity (K-rule) of propositional quantifier in your non-classical logic:
$$\begin{array}{c} \hline \vdash_M \forall P.(A \rightarrow B) \rightarrow (\forall P.A \rightarrow \forall P.B)\end{array}$$
and this introduction rule (N-rule):
$$\begin{array}{c} \vdash_M A \\ \hline \vdash_M \forall P.A\end{array}$$
The non-classical logic could be minimal logic, because in minimal logic this here is derivable:
$$ \vdash_M(Q \leftrightarrow (Q \rightarrow P)) \rightarrow P$$
The rest would then follow from N-rule and K-rule in the corresponding propositional quantified minimal logic.