understanding intuitionistic logic

Question

Just started a chapter on Intuitionism, and already I'm kinda confused on how to structure a proof in favor of intuitionistic logic.

And so I made an attempt at the two following practice exercises:

1. $\neg X \lor ¬Y \implies \neg(X \land Y)$

Attempt for 1: If we have a proof of the premise, then we have either a proof of $\neg X$ or a proof of $\neg Y$. Assume now that we are given a proof of $X$. It follows that our first proof could not have been a proof of $\neg X$: for that would be a proof that $X$ cannot be proved, and we are assuming that we have just been given a proof of $X$. So our first proof must in fact been a proof of $\neg X$ and the other a proof of $\neg Y$.

2. $\cfrac{\neg\exists x\neg F x \quad \forall x(Fx \lor \neg F x)}{\forall x Fx}$

Attempt for 2: Assume we are given a proof of its premise, then we have a proof that for some $n$, $\neg F n$ and for every $n$, we have either a proof of $Fx$ or $\neg Fx$. Using (9)(vi), we have a proof of a particular substitution, say $\neg Fk$. The question is whether there is an operation that transforms every proof of for all $x$, $F(x)$ into a proof of $0 = 1$. There is: given a proof of the universal quantification, apply it to generate a proof of $Fk$; this, together with the proof of $\neg Fk$, yields a proof of $0 = 1$.

All help is greatly appreciated!

Answer 1

The BHK (Brouwer-Heyting-Kolmogorov) interpretation (the one you describe) sees statements as proofs of these statement. So the way you need to reason is the following: If I want to show that from a statement $X$ I can conclude statement $Y$, then I need to show that if I have a proof of statement $X$ I can describe a proof for statement $Y$.

Let's start from an easy example. I want to show $P\to P\lor Q$. So I want to show that I can find a proof of $P\to P\lor Q$. By definition, I need to come up with an operation that takes a proof of $P$ and transforms it into a proof of $P\lor Q$. A proof of $P\lor Q$ is either a proof of $P$ or a proof of $Q$. So what would that operation be? In this case it's quite simple, the operation does nothing. It takes a proof of $P$ and gives me back the same proof. Now this proof of $P$ is also a proof of $P\lor Q$ (by the definition of the proof of $P\lor Q$).

Now let's look at the derivations that you have in your question. Assuming $\lnot X\lor\lnot Y$ I want to conclude $\lnot(X\land Y)$ (that is, assuming that I have a proof of $\lnot X\lor\lnot Y$) to find a proof of $\lnot(X\land Y)$).

So first of all think of what it is that you need to come up with: A proof for $\lnot(X\land Y)$ is by definition of proof of $X\land Y\to 0=1$, or (again using the definition) an operation that takes a proof of $X\land Y$ and gives a proof of $0=1$.

Now let's look at what we have. We have a proof of $\lnot X\lor\lnot Y$, that is either a proof of $\lnot X$ or a proof of $\lnot Y$.

Let's assume that we have a proof of $\lnot X$. This means that we have an operation $O_x$ that takes a proof of $X$ and gives us a proof of $0=1$. Can we use this operation to produce an operation that takes a proof of $X\land Y$ and produces a proof of $0=1$? Sure, a proof of $X\land Y$ is a proof of $X$ and a proof of $Y$. So if I have a proof of $X$ and a proof of $Y$ I can use operation $O_x$ on the proof of $X$ to find a proof $0=1$. What I just described is an operation that takes a proof of $X\land Y$ and gives a proof of $0=1$.

So now let's assume that we have a proof of $\lnot Y$. Can you come up with an operation that takes a proof of $X\land Y$ and gives you a proof of $0=1$?

The second example should be similar:

We assume that we have a proof of $\lnot\exists x\lnot Fx$ (what does that mean? try to break this down) and a proof of $\forall x(Fx\lor\lnot Fx)$ and we need to find a proof of $\forall x Fx$ (that is an operation that takes every element $n$ and and gives a proof of $Fn$.

I hope this will help you understand a little better the BHK interpretation. If things are unclear let me know and I will try to provide more details.