2.4 KiB
2.4 KiB
| title | date | tags | header-includes | ||
|---|---|---|---|---|---|
| undocumented idris2 features | 2022-11-12 |
|
<style> .sidebyside { display: grid; grid-template-columns: 1fr 1fr; } .sidebyside :is(.input, .goal) { box-sizing: border-box; width: 95%; height: 95%; } .sidebyside .input { grid-area: 0 / 1; } .sidebyside .goal { grid-area: 1 / 2; } </style> |
blah_rhs : w = w
``` {.idris .input}
blah2 : (n : Nat) -> 2 * n = n + (n + 0)
blah2 n with %syntactic (2 * n)
blah2 n | w = ?blah2_rhs
w : Nat
n : Nat
------------------------------
blah2_rhs : w = plus n (plus n 0)
:::
in blah2, only the exact syntactic occurrence of 2 * n is replaced, and
the n + (n + 0) is left alone.
equality proof in with
a with-abstraction can also have a proof of equality between the pattern
and the original expression, like the old inspect pattern.
:::sidebyside
blah : (n : Nat) -> 2 * n = n + (n + 0)
blah n with (2 * n) proof eq
blah n | w = ?blah_rhs
w : Nat
n : Nat
eq : plus n (plus n 0) = w
------------------------------
blah_rhs : w = w
:::