# Questions tagged [intuitionistic-logic]

Intuitionistic logic refer constructive logic, a logical system avoiding deduction rules like *Reductio ad absurdum*.

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### Superintuitionistic logics weaker then classical logic

I want to construct an example of a superintuitionistic logic which is a strict superset of $\operatorname{Int}$ and a strict subset of $\operatorname{Cl}$. I think that such logic can be obtained by ...
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### Constructive proof that any two natural numbers are either equal or not equal

I've been reading a bit about constructive mathematics and one of the first challenges I've found is to prove that any two natural numbers are either equal or not equal assuming only intuitionistic ...
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### Modal Heyting Algebras

Is there a standard way to add modal operators over a Heyting algebra -- as it was done e.g. by Johnstone and Tarski for Boolean algebras? Does this provide a semantics to some version of ...
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### There does not exist an even prime greater than two: is there an intuitionistic proof?

My own feeble understanding of intuitionism is that it is constructive and that, by not accepting the law of excluded middle, proofs by contradiction are eschewed. For my background on the topic, see ...
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### What is the algorithm for automated theorem proving in intuitionistic propositional logic?

In classical logic exists law of excluded middle: (a or not a). We can append not a to the knowledge base and show contradiction....
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### Can a proposition be false and its negation be false in Intuitionistic logic?

The principle of non-contradiction say that p and $\neg$ p can not be both true. If you consider the law of excluded middle as relevant, then either p or $\neg$ p is true, and thus the other is false ...
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### Equivalence of the Classical and Approximate Intermediate Value Theorems in Classical Logic

IVT (Classical) Let $f:\left[a,b\right]\to\mathbb{R}$ be continuous. If $f\left(a\right)<0$ and $f\left(b\right)>0$, then there exists $c\in\left(a,b\right)$ such that $f\left(c\right)=0$. In ...
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### How to prove that $\vdash \neg \neg (p \vee \neg p)$? (Natural Deduction) [closed]
I do know that double negation and LEM are equivalent, but can we prove $$\vdash \neg \neg (p \vee \neg p)$$ without using either of them, in a Fitch-style proof?