This commit is contained in:
Wen Kokke 2020-07-02 13:06:24 +01:00
parent ce40f1d96f
commit 6edc6e5b85

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@ -280,8 +280,8 @@ hold, then `m ≤ p` holds. Again, `m`, `n`, and `p` are implicit:
≤-trans (s≤s m≤n) (s≤s n≤p) = s≤s (≤-trans m≤n n≤p)
```
Here the proof is by induction on the _evidence_ that `m ≤ n`. In the
base case, the first inequality holds by `z≤n` and must show `zero ≤
p`, which follows immediately by `z≤n`. In this case, the fact that
base case, the first inequality holds by `z≤n` and must show `zero ≤ p`,
which follows immediately by `z≤n`. In this case, the fact that
`n ≤ p` is irrelevant, and we write `_` as the pattern to indicate
that the corresponding evidence is unused.