Make sure we do not remove the outer `x: 5` (line 6), even though it
is a tie with the inner `x: 5` (and the inner `x: 5` occurs earlier in
the file (line 2) so it would win in a tie-break).

What happens here is that we establish that line 1 requires (depends
on) line 6. Even though within x we have a tie between line 2 and line
6, we find that line 1 cannot be removed (because y only has 1 winner
in line 4, and line 4 requires line 1), and so therefore line 6 is
required, and so line 2 is redundant.

Contrast with 57.

-- a.cue --
if x == 5 {
	x: 5
	x: int
	y: "hi"
}
x: 5
-- out/trim-v3 --
== a.cue
if x == 5 {
	y: "hi"
}
x: 5
-- out/trim-v3-noshare --
== a.cue
if x == 5 {
	y: "hi"
}
x: 5
