# SE-0309: Unlock existential types for all protocols **URL:** \ **Category:** Proposal Reviews\ **Created:** [April 17, 2021, 7:56pm UTC](https://forums.swift.org/t/se-0309-unlock-existential-types-for-all-protocols/47515 "2021-04-17T19:56:29Z")\ **Posts on this page:** 1\ **Showing post:** 122
**Author:** ![Dante-Broggi](https://sea2.discourse-cdn.com/swift/user_avatar/forums.swift.org/dante-broggi/32/7955_2.png) [@Dante-Broggi](https://forums.swift.org/u/Dante-Broggi)\ **Post date:** [May 2, 2021, 5:23pm UTC](https://forums.swift.org/t/se-0309-unlock-existential-types-for-all-protocols/47515/122 "2021-05-02T17:23:45Z")
> [@sighoya](#): > > I disagree. But maybe the term `existential` is redefined in Swift context, but this would lead to more confusion for users coming from other languages already knowing existentials. So, I believe that this side discussion is fundamentally about a disagreement over the definition of "existential". IIUC, your definition of "existential" is similar to: > a value of an existential of protocol `P` is a value `v` for which there exists a type `T`, such that `v` is of type `T` and `T` conforms to `P`. Under this definition the value boxed in `any P` and both possible extensions of `some P` are existentials. * * * However, IIUC the definition of "existential" in general use in the Swift community is similar to: > a value of an existential of protocol `P` is a value **`b` which contains a value** `v` for which there exists a type `T`, such that `v` is of type `T` and `T` conforms to `P`, **and `b` contains proofs of those statements about `v`**. Under this definition, `any P`, but not the value boxed in it, is an existential, but `some P` is not, because the proofs for `some P` are not stored in the value, but in the context. --- _[View the full topic](https://forums.swift.org/t/se-0309-unlock-existential-types-for-all-protocols/47515)._