$(p \to (q \to r)) \to ((p \to r) \lor (q \to r))$
$(p \to (q \lor r)) \to ((p \to q) \lor (p \to r))$
are holded in intuitionistic logics?
I could not found the model of intuitional logic that aboves are not true.
So, if these can be proved, give me a proof of it. Unless, I wanna get the inconsistent model of intuitionistic logics with those two sentences.
Welcome to MSE!
Hint:
The first claim is not even true classically. Can you find a way to falsify it in a very simple boolean algebra? (For instance, one with $4$ elements?)
The second claim is true classically (replace $a \to b$ by $\lnot a \lor b$ everywhere to see why), but is not true intuitionistically. To see why, you'll want a heyting algebra which falsifies it. And it's not worth looking at boolean ones, since we know the claim is true classically.
Since you have $3$ primitive propositions ($p$, $q$, and $r$), you might try to find a model where $p$, $q$, and $r$ are all incomparable. I'll tell you that there's a heyting algebra with $9$ elements that does the trick. Can you take it from here?
Edit:
After your edit, the first claim is true classically (as you've noted), but is still false intuitionistically. To see why, consider the same $9$ element heyting algebra that I alluded to for your second proposition. Again, you should put $p$, $q$, and $r$ along the middle row of this lattice. Can you compute $q \to r$, $p \to q \to r$, etc. and see that this falsifies the claim?
As a bonus hint, while you put $p$, $q$, and $r$ in the middle both times, you'll have to put them in different positions for proposition $1$ and $2$. Thankfully there's only $3$ permutations to try, so it shouldn't take you long to check them all.
I hope this helps ^_^