1a77f8b63d
If we have an inequality key that constrains the other end of the index, it doesn't directly help us in doing the initial positioning ... but it does imply a NOT NULL constraint on the index column. If the index stores nulls at this end, we can use the implied NOT NULL condition for initial positioning, just as if it had been stated explicitly. This avoids wasting time when there are a lot of nulls in the column. This is the reverse of the examples given in bugs #6278 and #6283, which were about failing to stop early when we encounter nulls at the end of the indexscan.
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