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-rw-r--r--prolog/problems/lists/conc_3/en.py2
1 files changed, 1 insertions, 1 deletions
diff --git a/prolog/problems/lists/conc_3/en.py b/prolog/problems/lists/conc_3/en.py
index 2e74e5e..c384222 100644
--- a/prolog/problems/lists/conc_3/en.py
+++ b/prolog/problems/lists/conc_3/en.py
@@ -30,7 +30,7 @@ hint = {
'eq_instead_of_equ': '''\
<p>The operator <code>==</code> is "stricter" than operator <code>=</code> in the sense that
for the latter it is enough to be able to make the two operands equal (unification). Perhaps by using <code>=</code>
-you can make the predicate <code>rev/2</code> more general (e.g. able to work with output arguments becoming inputs).</p>
+you can make the predicate <code>conc/3</code> more general (e.g. able to work with output arguments becoming inputs).</p>
<p>Of course, you can also solve the exercise without explicit use of either of these two operators, just
remember that unification is implicitly performed with the predicate's arguments (head of clause).</p>
''',