How does Equatable synthesis interact with implementing refined protocols?

Using Swift
6

That's an interesting point and potentially counterintuitive. Conforming to two protocols with default implementations would ordinarily require the concrete type to implement the method; in this case, you do get synthesis back because of that.

Therefore, you could work around your issue by creating an internal protocol refining Equatable named _RestoreSynthesizedEquatableConformance with a default implementation of ==:

protocol _RestoreSynthesizedEquatableConformance : Equatable { }

extension _RestoreSynthesizedEquatableConformance {
  static func == (lhs: Self, rhs: Self) -> Bool { fatalError() }
}

struct S { }

extension S : Strideable, _RestoreSynthesizedEquatableConformance {
  func advanced(by n: Int) -> S { return S() }
  func distance(to other: S) -> Int { return 0 }
}

S() == S() // true
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