Distribution of disjunction and conjunction over each other in intuitionistic logic

Question

I can't find any references as to whether or not the usual properties of disjunction and conjunction distributing over each other hold in intuitionistic logic. Consider:

$$(1) \ \ (p \vee (q \& r)) \rightarrow ((p \vee q) \& (p \vee r))$$
$$(2) \ \ ((p \& q) \vee (p \& r)) \rightarrow (p \& (q \vee r))$$
$$(3) \ \ ((p \vee q) \& (p \vee r)) \rightarrow (p \vee (q \& r))$$
$$(4) \ \ (p \& (q \vee r)) \rightarrow ((p \& q) \vee (p \& r))$$

I can easily prove (1) and (2) in intuitionistic logic. But (3) and (4) do not seem provable. Are they intuitionistically valid?

Answer 1

Here are proofs in a fairly standard natural deduction system for intuitionistic logic.

For (3), assume $(p\lor q)\&(p\lor r)$. In particular, we have $p\lor q$, so we can consider separately two cases.

Case 1: $p$. Then the desired conclusion $p\lor(q\&r)$ follows immediately.

Case 2: $q$. In this case, we go back to the initial assumption to obtain $p\lor r$. So we can consider two subcases.

Subcase 2a: $p$. Again the desired conclusion follows immediately.

Subcase 2b: $r$. Combine this with $q$, which we have because we're in Case 2, to get $q\&r$ and thus the desired conclusion $p\lor(q\&r)$.

We got the desired conclusion in all cases and subcases, so the proof is complete.

For (4), assume $p\&(q\lor r)$. So we know $p$ and we know $q\lor r$. The latter lets us consider two cases separately.

Case 1: $q$. So we have $p\&q$, and therefore $(p\&q)\lor(p\&r)$, as required.

Case 2: $r$. So we have $p\&r$, and therefore $(p\&q)\lor(p\&r)$, as required.