diff --git a/src/backend/nodes/bitmapset.c b/src/backend/nodes/bitmapset.c
index 9cda3b1cc1..2ee7385f02 100644
--- a/src/backend/nodes/bitmapset.c
+++ b/src/backend/nodes/bitmapset.c
@@ -9,12 +9,7 @@
  * the minimum possible number of words, i.e, there are never any trailing
  * zero words.  Enforcing this requires that an empty set is represented as
  * NULL.  Because an empty Bitmapset is represented as NULL, a non-NULL
- * Bitmapset always has at least 1 Bitmapword.  We can exploit this fact to
- * speedup various loops over the Bitmapset's words array by using "do while"
- * loops instead of "for" loops.  This means the code does not waste time
- * checking the loop condition before the first iteration.  For Bitmapsets
- * containing only a single word (likely the majority of them) this reduces
- * the loop condition tests by half.
+ * Bitmapset always has at least 1 Bitmapword.
  *
  *
  * Copyright (c) 2003-2023, PostgreSQL Global Development Group
@@ -96,8 +91,6 @@ bms_copy(const Bitmapset *a)
 bool
 bms_equal(const Bitmapset *a, const Bitmapset *b)
 {
-	int			i;
-
 	/* Handle cases where either input is NULL */
 	if (a == NULL)
 	{
@@ -108,17 +101,20 @@ bms_equal(const Bitmapset *a, const Bitmapset *b)
 	else if (b == NULL)
 		return false;
 
-	/* can't be equal if the word counts don't match */
-	if (a->nwords != b->nwords)
+	/*
+	 * Most Bitmapsets will likely just have 1 word, so check for equality
+	 * there before checking the number of words match.  The sets cannot be
+	 * equal when they don't have the same number of words.
+	 */
+	if (a->words[0] != b->words[0] || a->nwords != b->nwords)
 		return false;
 
-	/* check each word matches */
-	i = 0;
-	do
+	/* check all the remaining words match */
+	for (int i = 1; i < a->nwords; i++)
 	{
 		if (a->words[i] != b->words[i])
 			return false;
-	} while (++i < a->nwords);
+	}
 
 	return true;
 }
@@ -353,25 +349,27 @@ bms_difference(const Bitmapset *a, const Bitmapset *b)
 bool
 bms_is_subset(const Bitmapset *a, const Bitmapset *b)
 {
-	int			i;
-
 	/* Handle cases where either input is NULL */
 	if (a == NULL)
 		return true;			/* empty set is a subset of anything */
 	if (b == NULL)
 		return false;
 
-	/* 'a' can't be a subset of 'b' if it contains more words */
-	if (a->nwords > b->nwords)
+	/*
+	 * Check the 0th word first as many sets will only have 1 word.
+	 * Otherwise, 'a' can't be a subset of 'b' if it contains more words, so
+	 * there's no need to check remaining words if 'a' contains more words
+	 * than 'b'.
+	 */
+	if ((a->words[0] & ~b->words[0]) != 0 || a->nwords > b->nwords)
 		return false;
 
-	/* Check all 'a' members are set in 'b' */
-	i = 0;
-	do
+	/* Check all remaining words to see 'b' has members not set in 'a'. */
+	for (int i = 1; i < a->nwords; i++)
 	{
 		if ((a->words[i] & ~b->words[i]) != 0)
 			return false;
-	} while (++i < a->nwords);
+	}
 	return true;
 }