Proof of contradiction vs refutation by contradiction and what is/isn't allowed in constructive logic?

Question

Here it is stated that constructive logic allows refutation by contradiction:

1. The proposition to be proved is ¬P.
2. Assume P.
3. Derive falsehood.
4. Conclude ¬P.

But not indirect proof:

1. The proposition to be proved is P.
2. Assume ¬P.
3. Derive falsehood.
4. Conclude P.

My question is what about the following case:

0. We have a P and ¬P where (P∧¬P) is logical inconsistency.
1. The proposition to be proved is ¬P.
2. Assume ¬¬P.
3. Derive falsehood.
4. Conclude ¬P.

Thinking about it, I think that the above can be rephrased as this:

0. We have a P and ¬P where (P∧¬P) is logical inconsistency.
1. The proposition to be proved is ¬P.
2. Assume ¬¬P.
3. Derive falsehood.
4. Conclude ¬¬¬P.

And if this is to be believed then this is perfectly fine in constructive logic and does not use indirect proof.

Can anyone say for sure if this is allowed in constructive logic or is it a form of indirect proof? How to make sense of this in a way other than symbol manipulation?

Answer 1

The sequence

2. Assume $\lnot \lnot P$.
3. Derive falsehood.
4. Conclude $\lnot P$.

doesn't work without modification. Deriving a contradiction from an assumption $A$ allows you to conclude $\lnot A$, so in this instance, you'd conclude $\lnot \lnot \lnot P$.

However, it's possible to eliminate a triple negation into a single negation. So suppose $\lnot \lnot \lnot P$ holds and we want to prove $\lnot P$. As you point out in your question, you can prove $\lnot P$ by assuming $P$ and deriving falsehood, so let's do that. This proof uses this same idea multiple times, so be sure to keep track of which assumptions are still active. Right now, we just have $\lnot \lnot \lnot P$ and $P$ among the hypotheses.

Knowing that $P$ is true, we can derive $\lnot \lnot P$ using an assumption that $\lnot P$ is true and getting a contradiction with the assumption of $P$. But then we have both $\lnot \lnot \lnot P$ and $\lnot \lnot P$, which is a contradiction.

So the first assumption that $P$ is true must be false, so we have $\lnot P$.

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By the way, it's worth noting that intuitionistic logic is not defined by what isn't allowed, but by what is allowed. Each connective (e.g. and, or, not) has introduction and elimination rules. Introduction rules tell you how to prove something with a specific type of conclusion and elimination rules tell you how you can use a hypothesis of a certain type.

For negation, the introduction rule for negation is that you can prove $\lnot P$ by assuming $P$ and deriving a contradiction. The elimination rule is that you can conclude a contradiction given $\lnot P$ and $P$. These rules are special cases of the rules for $A \to B$ with $\lnot P := P \to \bot$, where $\bot$ is the always false proposition.