TITLE: distribution of categorical product (conjunction) over coproduct (disjunction)
QUESTION [6 upvotes]: In the classical and intuitionistic propositional calculi, it is straightforward, using natural deduction, to derive $((A \land C) \lor (B \land C))$ from $(A \lor B) \land C$:


Assume $(A \lor B) \land C$.


$A \lor B$ by conjunction elimination with 1.


$C$ by conjunction elimination with 1.



Assume $A$.




$A \land C$ by conjunction introduction with 3 and 4.




$(A \land C) \lor (B \land C)$ by disjunction introduction with 5.




Assume $B$.




$B \land C$ by conjunction introduction with 3 and 7.




$(A \land C) \lor (B \land C)$ by disjunction introduction with 8.



$(A \land C) \lor (B \land C)$ by disjunction elimination with 2, 4–6, and 7–9.


In a categorical treatment of the same, conjunction is a product, disjunction is a coproduct, and the task is to find an arrow $h\colon (A \lor B) \land C \to (A \land C) \lor (B \land C)$.  I've been drawing commutative diagrams for a few hours yet, and no such arrow is presenting itself.  The logic has products, coproducts, and exponentials, and should be, as I understand it, bicartesian closed.  That definition of bicartesian closed includes the condition that products distribute over coproducts, and adds an appropriate equation.  Do I have to appeal to that equation to get $h$, or can I demonstrate the arrow I want without it?
As to motivation, I've already got arrows
$f\colon A \land C \to D$
and
$g\colon B \land C \to D$,
so I can construct the coproduct arrow 
$[f,g]\colon (A \land C) \lor (B \land C) \to D$.
Had I $h$, I could construct what I really want:
$h \circ [f,g]\colon (A \lor B) \land C \to D$.
If I can't construct an arrow $h$, as described, I'd still be fine with a way of demonstrating, given an arrow $(A \lor B) \land C \to D$, another arrow $(A \land C) \lor (B \land C) \to D$ that doesn't appeal to an arrow like $h$.

REPLY [5 votes]: You can curry $f$ and $g$ to get arrows $f' \colon A \to D^C$ and $g' \colon B \to D^C$. The copairing of those is then $[f',g']\colon (A \lor B) \to D^C$, which can then be uncurried to get an arrow $(A \lor B) \land C \to D$. This approach can also be used to construct the distribution arrow, if you want that instead.
Note that this only relies on conjunction providing a monoidal closed category, and does not require that conjunction be the categorical product. A dual construction is possible in a co-closed (is that the right term?) category using co-currying to distribute over a product, such as in linear logic where multiplicative disjunction distributes over additive conjunction.