I asked a question here about whether $$ [ (P \wedge \neg R) \to Q ] \to [ P \to (Q \vee R) ] $$ is a tautology in intuitionistic math. The answer is "no". What prompted my question was my proof of the following: Let $a,b,n \in \mathbb{N}$. If $n \neq 1$ and $n$ divides both $a$ and $b$, then $b$ is a composite number or $b$ divides $a$. My proof:
Suppose $b$ is not composite. Then $b$ is prime. Since $n \neq 1$ and $n$ divides $b$, we must have $n = b$. Thus $b$ divides $a$.
This proof has the form $(P \wedge \neg R) \to Q$. If we were using normal first order logic, I could conclude $P \to (Q \vee R)$. But, using intuitionistic logic, I cannot conclude that which was to be proven. How would you carry out this proof in intuitionistic logic?