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+A parameter type is missing an explicit lifetime bound and may not live long
+enough.
+
+Erroneous code example:
+
+```compile_fail,E0309
+// This won't compile because the applicable impl of
+// `SomeTrait` (below) requires that `T: 'a`, but the struct does
+// not have a matching where-clause.
+struct Foo<'a, T> {
+    foo: <T as SomeTrait<'a>>::Output,
+}
+
+trait SomeTrait<'a> {
+    type Output;
+}
+
+impl<'a, T> SomeTrait<'a> for T
+where
+    T: 'a,
+{
+    type Output = u32;
+}
+```
+
+The type definition contains some field whose type requires an outlives
+annotation. Outlives annotations (e.g., `T: 'a`) are used to guarantee that all
+the data in `T` is valid for at least the lifetime `'a`. This scenario most
+commonly arises when the type contains an associated type reference like
+`<T as SomeTrait<'a>>::Output`, as shown in the previous code.
+
+There, the where clause `T: 'a` that appears on the impl is not known to be
+satisfied on the struct. To make this example compile, you have to add a
+where-clause like `T: 'a` to the struct definition:
+
+```
+struct Foo<'a, T>
+where
+    T: 'a,
+{
+    foo: <T as SomeTrait<'a>>::Output
+}
+
+trait SomeTrait<'a> {
+    type Output;
+}
+
+impl<'a, T> SomeTrait<'a> for T
+where
+    T: 'a,
+{
+    type Output = u32;
+}
+```