diff --git a/Documentation/SBV/Examples/KnuckleDragger/Lists.hs b/Documentation/SBV/Examples/KnuckleDragger/Lists.hs
index fce9c2a6..d49f2487 100644
--- a/Documentation/SBV/Examples/KnuckleDragger/Lists.hs
+++ b/Documentation/SBV/Examples/KnuckleDragger/Lists.hs
@@ -11,6 +11,10 @@
 
 {-# LANGUAGE CPP                 #-}
 {-# LANGUAGE DataKinds           #-}
+{-# LANGUAGE DeriveAnyClass      #-}
+{-# LANGUAGE DeriveDataTypeable  #-}
+{-# LANGUAGE StandaloneDeriving  #-}
+{-# LANGUAGE TemplateHaskell     #-}
 {-# LANGUAGE TypeAbstractions    #-}
 
 {-# OPTIONS_GHC -Wall -Werror -Wno-unused-do-bind #-}
@@ -18,14 +22,17 @@
 module Documentation.SBV.Examples.KnuckleDragger.Lists where
 
 import Data.SBV
-import qualified Data.SBV.List as L
+import Data.SBV.List
 import Data.SBV.Tools.KnuckleDragger
 
+import Prelude hiding(reverse, (++), any, all, notElem, filter, map)
+
 #ifndef HADDOCK
 -- $setup
 -- >>> -- For doctest purposes only:
 -- >>> :set -XScopedTypeVariables
 -- >>> import Control.Exception
+-- >>> import Data.SBV.Tools.KnuckleDragger
 #endif
 
 -- | A list of booleans is not all true, if any of them is false. We have:
@@ -35,7 +42,7 @@ import Data.SBV.Tools.KnuckleDragger
 -- [Proven] allAny
 allAny :: IO Proof
 allAny = runKD $ lemma "allAny" (\(Forall @"xs" xs) -> p xs) [induct p]
-  where p xs = sNot (L.all id xs) .== L.any sNot xs
+  where p xs = sNot (all id xs) .== any sNot xs
 
 -- | If an integer list doesn't have 2 as an element, then filtering for @> 2@ or @.>= 2@
 -- yields the same result. We have:
@@ -45,7 +52,7 @@ allAny = runKD $ lemma "allAny" (\(Forall @"xs" xs) -> p xs) [induct p]
 -- [Proven] filterEx
 filterEx :: IO Proof
 filterEx = runKD $ lemma "filterEx" (\(Forall @"xs" xs) -> p xs) [induct p]
-  where p xs = (2 :: SInteger) `L.notElem` xs .=> (L.filter (.> 2) xs .== L.filter (.>= 2) xs)
+  where p xs = (2 :: SInteger) `notElem` xs .=> (filter (.> 2) xs .== filter (.>= 2) xs)
 
 -- | The 'filterEx' example above, except we get a counter-example if `2` can be in the list. Note that
 -- we don't even need the induction tactic here. (Though having it wouldn't hurt.) We have:
@@ -58,8 +65,79 @@ filterEx = runKD $ lemma "filterEx" (\(Forall @"xs" xs) -> p xs) [induct p]
 filterEx2 :: IO ()
 filterEx2 = runKD $ do
         let p :: SList Integer -> SBool
-            p xs = L.filter (.> 2) xs .== L.filter (.>= 2) xs
+            p xs = filter (.> 2) xs .== filter (.>= 2) xs
 
         lemma "filterEx" (\(Forall @"xs" xs) -> p xs) []
 
         pure ()
+
+data A
+mkUninterpretedSort ''A
+
+data B
+mkUninterpretedSort ''B
+
+-- | An uninterpreted function, for use as a witness in the following higher-order proofs.
+f :: SA -> SB
+f = uninterpret "f"
+
+-- | @reverse (x:xs) == reverse xs ++ [x]@
+--
+-- >>> runKD revCons
+-- Lemma: revCons                          Q.E.D.
+-- [Proven] revCons
+revCons :: KD Proof
+revCons = do
+    let p :: SA -> SList A -> SBool
+        p x xs = reverse (x .: xs) .== reverse xs ++ singleton x
+
+    lemma "revCons" (\(Forall @"x" x) (Forall @"xs" xs) -> p x xs) []
+
+-- | @map f (xs ++ ys) == map f xs ++ map f ys@
+--
+-- >>> runKD mapAppend
+-- Lemma: mapAppend                        Q.E.D.
+-- [Proven] mapAppend
+mapAppend :: KD Proof
+mapAppend = do
+   let p :: (SA -> SB) -> SList A -> SList A -> SBool
+       p g xs ys = map g (xs ++ ys) .== map g xs ++ map g ys
+
+   lemma "mapAppend"
+         (\(Forall @"xs" xs) (Forall @"ys" ys) -> p f xs ys)
+         -- induction is done on the last argument, so flip to do it on xs
+         [induct (flip (p f))]
+
+-- | @map f . reverse == reverse . map f@
+--
+-- >>> runKD mapReverse
+-- Lemma: revCons                          Q.E.D.
+-- Lemma: mapAppend                        Q.E.D.
+-- Chain: mapReverse
+--   Lemma: mapReverse.1                   Q.E.D.
+--   Lemma: mapReverse.2                   Q.E.D.
+--   Lemma: mapReverse.3                   Q.E.D.
+--   Lemma: mapReverse.4                   Q.E.D.
+--   Lemma: mapReverse.5                   Q.E.D.
+--   Lemma: mapReverse.6                   Q.E.D.
+-- Lemma: mapReverse                       Q.E.D.
+-- [Proven] mapReverse
+mapReverse :: KD Proof
+mapReverse = do
+     let p :: (SA -> SB) -> SList A -> SBool
+         p g xs = reverse (map g xs) .== map g (reverse xs)
+
+     rCons <- revCons
+     mApp  <- mapAppend
+
+     chainLemma "mapReverse"
+                (\(Forall @"xs" xs) -> p f xs)
+                (\x xs -> [ reverse (map f (x .: xs))
+                          , reverse (f x .: map f xs)
+                          , reverse (map f xs) ++ singleton (f x)
+                          , map f (reverse xs) ++ singleton (f x)
+                          , map f (reverse xs) ++ map f (singleton x)
+                          , map f (reverse xs ++ singleton x)
+                          , map f (reverse (x .: xs))
+                          ])
+                [induct (p f), rCons, mApp]