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.

1. If you consider the law of excluded middle as relevant, then either p or $$\neg$$ p is true, and thus the other is false by the principle of non-contradiction.
2. But if you consider the excluded middle as not relevant, just like intuitionistic logic, what would avoid some p and $$\neg$$ p to be both false ?

Actually, if true is the same than provable, then such a proposition is just a Gödel one : p is not provable and neither $$\neg$$ p is.

I think this is a very good interpretation and a big demystification of Gödel theorem : it is not about some true that we can not reach, it is about some absurdity that is still an absurdity when you negate it.

• It is not possible, if by "$p$ is false" you mean "$\lnot p$ is provable". Indeed, then "$p$ and $\lnot p$ are both false" translate to "$\lnot p$ and $\lnot \lnot p$ are both provable", which is a contradiction. But "$p$ is false" is different from "$p$ is not provable". Mar 30 '21 at 14:13
• No; the derivation of $\bot$ from $A \to \bot$ and $(A \to \bot) \to \bot$ is intuitionistically valid. Mar 30 '21 at 14:15
• About the purported "big demystification of Gödel theorem" we have to note that G's Th is sound from an intuitionistic point of view... but it is "obvious". It is a cornerstone of Brouwer's view that Hilbert's assumption that every mathematical problem is decidable was not tenable. Mar 30 '21 at 14:17
• I want to define a recursive logic where false is no longer define as $\vDash \neg p$ but as $\nvDash p$. Mar 31 '21 at 9:14

The statement "$$p$$ and $$\neg p$$ are both false" is formally expressed as $$\neg p \land \neg \neg p,$$ which is false by the law of non-contradiction, which states that for any $$q$$ the proposition $$q \land \neg q$$ is false (take $$q = \neg p$$).
To put it another way, it is very easy to find a formula which is neither provable nor falsifiable, if we allow free variables, for example $$x < 42$$ where $$x$$ ranges over the natural numbers (say, in Peano arithmetic) is not provable, but neither is $$\neg (x < 42).$$