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$$ \dfrac{\dfrac{\dfrac{\dfrac{}{A\vdash A}}{A\vdash A\lor(A\to C)}\qquad\dfrac{}{C\vdash C}}{\dfrac{\dfrac{(A\lor(A\to C))\to C,A\vdash C}{(A\lor(A\to C))\to C\vdash A\to C}}{(A\lor(A\to C))\to C\vda …

Henkin-style completeness proofs for intuitionistic logic are perfectly possible: instead of maximal consistent sets, you consider, for each formula $A$, maximal sets $\Gamma$ such that $\Gamma\nvdash …

No, every consistent propositional logic that extends intuitionistic logic is a sublogic of classical logic. (That’s why consistent superintuitionistic logics are also called intermediate logics.) To …

A related property in which the implicational fragment differs from full intuitionistic logic is that the former is locally finite: for every finite $n$, there are only finitely many inequivalent form …

Let $F,G$ be the two frames. Let $\beta$ be the frame formula of $F$ (using notation from the Chagrov and Zakharyaschev book you mention in the MSE question, $\beta=\beta^\sharp(F,\bot)$). Since $\bet …

$\let\eq\leftrightarrow$Notice that $\psi(A)=u$ iff $\vdash_\mathrm{CPC}A\eq u$ iff $\vdash_\mathrm{IPC}\neg\neg(A\eq u)$. (I will write just $\vdash$ for $\vdash_\mathrm{IPC}$.) Thus: $\bot$ has a …

The finite model property of intuitionistic logic implies that every unprovable formula has finite rank. On the other hand, all positive integers are ranks of some formulas; there are many families of …

Both are false. Consider the following Kripke model $M\vDash Q^e$ (in fact, it satisfies the intuitionistic version of $\mathrm{PA}^-$): it consists of two worlds $u,v$ such that $u$ sees $v$; the fir …

The result is actually false, for $m=6$. (One can bring it down to $m=2$ with a bit of effort.) Let $n$ be arbitrarily large, and $\phi_n(\vec q)$ be a Jankov–De Jongh frame formula of $\def\p#1{\lang …