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Answers and rating deltas for
Why does `forall (a :: j) (b:: k)` work differently than `forall (p :: (j,k))`?
| Author | Votes | Δ |
|---|---|---|
| András Kovács | 3 | -0.09 |
| luqui | 3 | +0.09 |
Last visited: Mar 15, 2019, 5:06:24 AM