`x` and `y` cannot be eliminated. This also means uses of `x` and `y`
can't be eliminated. So the simplification of `z` can only remove the
literal `{a: 5}`.

Similar vein to 5.

-- a.cue --
x: {a: 5}
y: {a: 5}
z: x & y & {a: 5}
-- out/trim-v3 --
== a.cue
x: {a: 5}
y: {a: 5}
z: x & y & _
-- diff/-out/trim-v3<==>+out/trim --
diff old new
--- old
+++ new
@@ -1,4 +1,4 @@
 == a.cue
 x: {a: 5}
 y: {a: 5}
-z: x & y & {a: 5}
+z: x & y & _
-- out/trim-v3-noshare --
== a.cue
x: {a: 5}
y: {a: 5}
z: x & y & _
-- diff/-out/trim-v3-noshare<==>+out/trim --
diff old new
--- old
+++ new
@@ -1,4 +1,4 @@
 == a.cue
 x: {a: 5}
 y: {a: 5}
-z: x & y & {a: 5}
+z: x & y & _
-- out/trim --
== a.cue
x: {a: 5}
y: {a: 5}
z: x & y & {a: 5}
