Function process_duplicate_ors

src/backend/optimizer/prep/prepqual.c:516–676

* process_duplicate_ors * Given a list of exprs which are ORed together, try to apply * the inverse OR distributive law. * * Returns the resulting expression (could be an AND clause, an OR * clause, or maybe even a single subexpression). */

 * clause, or maybe even a single subexpression).
 */
static Expr *
process_duplicate_ors(List *orlist)
{
	List	   *reference = NIL;
	int			num_subclauses = 0;
	List	   *winners;
	List	   *neworlist;
	ListCell   *temp;

	/* OR of no inputs reduces to FALSE */
	if (orlist == NIL)
		return (Expr *) makeBoolConst(false, false);

	/* Single-expression OR just reduces to that expression */
	if (list_length(orlist) == 1)
		return (Expr *) linitial(orlist);

	/*
	 * Choose the shortest AND clause as the reference list --- obviously, any
	 * subclause not in this clause isn't in all the clauses. If we find a
	 * clause that's not an AND, we can treat it as a one-element AND clause,
	 * which necessarily wins as shortest.
	 */
	foreach(temp, orlist)
	{
		Expr	   *clause = (Expr *) lfirst(temp);

		if (is_andclause(clause))
		{
			List	   *subclauses = ((BoolExpr *) clause)->args;
			int			nclauses = list_length(subclauses);

			if (reference == NIL || nclauses < num_subclauses)
			{
				reference = subclauses;
				num_subclauses = nclauses;
			}
		}
		else
		{
			reference = list_make1(clause);
			break;
		}
	}

	/*
	 * Just in case, eliminate any duplicates in the reference list.
	 */
	reference = list_union(NIL, reference);

	/*
	 * Check each element of the reference list to see if it's in all the OR
	 * clauses.  Build a new list of winning clauses.
	 */
	winners = NIL;
	foreach(temp, reference)
	{
		Expr	   *refclause = (Expr *) lfirst(temp);