diff --git a/chia/_tests/core/mempool/test_mempool_manager.py b/chia/_tests/core/mempool/test_mempool_manager.py
index 0db312df78b1..45c80258a2a0 100644
--- a/chia/_tests/core/mempool/test_mempool_manager.py
+++ b/chia/_tests/core/mempool/test_mempool_manager.py
@@ -770,7 +770,16 @@ def make_test_coins() -> list[Coin]:
     return ret
 
 
+def make_ephemeral(coins: list[Coin]) -> list[Coin]:
+    ret: list[Coin] = []
+    for i, parent in enumerate(coins):
+        ret.append(Coin(parent.name(), height_hash(i + 150), uint64(i * 100)))
+    return ret
+
+
 coins = make_test_coins()
+eph = make_ephemeral(coins)
+eph2 = make_ephemeral(eph)
 
 
 @pytest.mark.parametrize(
@@ -800,6 +809,26 @@ def make_test_coins() -> list[Coin]:
         ([mk_item(coins[0:2])], mk_item(coins[0:2], fee=10000000), True),
         # or if we spend the same coins with additional coins
         ([mk_item(coins[0:2])], mk_item(coins[0:3], fee=10000000), True),
+        # SUPERSET RULE WITH EPHEMERAL COINS
+        # the super set rule only takes non-ephemeral coins into account. The
+        # ephmeral coins depend on how we spend, and might prevent legitimate
+        # replace-by-fee attempts.
+        # replace a spend that includes an ephemeral coin with one that doesn't
+        ([mk_item(coins[0:2] + eph[0:1])], mk_item(coins[0:2], fee=10000000), True),
+        # replace a spend with two-levels of ephemeral coins, with one that
+        # only has 1-level
+        ([mk_item(coins[0:2] + eph[0:1] + eph2[0:1])], mk_item(coins[0:2] + eph[0:1], fee=10000000), True),
+        # replace a spend with two-levels of ephemeral coins, with one that
+        # doesn't
+        ([mk_item(coins[0:2] + eph[0:1] + eph2[0:1])], mk_item(coins[0:2], fee=10000000), True),
+        # replace a spend with two-levels of ephemeral coins, with one that
+        # has *different* ephemeral coins
+        ([mk_item(coins[0:2] + eph[0:1] + eph2[0:1])], mk_item(coins[0:2] + eph[1:2] + eph2[1:2], fee=10000000), True),
+        # it's OK to add new ephemeral spends
+        ([mk_item(coins[0:2])], mk_item(coins[0:2] + eph[1:2] + eph2[1:2], fee=10000000), True),
+        # eph2[0:1] is not an ephemeral coin here, this violates the superset
+        # rule. eph[0:1] is missing for that
+        ([mk_item(coins[0:2] + eph2[0:1])], mk_item(coins[0:2] + eph[1:2] + eph2[1:2], fee=10000000), False),
         # FEE- AND FEE RATE RULES
         # if we're replacing two items, each paying a fee of 100, we need to
         # spend (at least) the same coins and pay at least 10000000 higher fee
diff --git a/chia/full_node/mempool_manager.py b/chia/full_node/mempool_manager.py
index 6a627c08cb94..ddb7c9e315ee 100644
--- a/chia/full_node/mempool_manager.py
+++ b/chia/full_node/mempool_manager.py
@@ -455,7 +455,7 @@ async def validate_spend_bundle(
             )
 
         if removal_names != removal_names_from_coin_spends:
-            # If you reach here it's probably because your program reveal doesn't match the coin's puzzle hash
+            # If you reach here it's probably because your puzzle reveal doesn't match the coin's puzzle hash
             return Err.INVALID_SPEND_BUNDLE, None, []
 
         removal_record_dict: dict[bytes32, CoinRecord] = {}
@@ -806,8 +806,13 @@ def can_replace(
         # bundle with AB with a higher fee. An attacker then replaces the bundle with just B with a higher
         # fee than AB therefore kicking out A altogether. The better way to solve this would be to keep a cache
         # of booted transactions like A, and retry them after they get removed from mempool due to a conflict.
-        for coin in item.removals:
-            if coin.name() not in removal_names:
+        conflicting_removals = {c.name(): c for c in item.removals}
+        for coin in conflicting_removals.values():
+            coin_name = coin.name()
+            # if the parent of this coin is one of the spends in this
+            # transaction, it means it's an ephemeral coin spend. Such spends
+            # are not considered by the superset rule
+            if coin_name not in removal_names and coin.parent_coin_info not in conflicting_removals:
                 log.debug(f"Rejecting conflicting tx as it does not spend conflicting coin {coin.name()}")
                 return False