RxSwift Observable<(T, T)>.distinctUntilChanged() returns wrong result

distinctUntilChanged.swift

let source = Observable<(Int, Int)>.from([
  (0, 1),
  (2, 3),
  (4, 5),
])

// Doesn't work as expected
// A -> (0, 1)
source
  .distinctUntilChanged(==)
  .subscribe(onNext: { print("A ->", $0) })
print()

// Doesn't work as expected
// B -> (0, 1)
source
  .distinctUntilChanged { $0 == $1 }
  .subscribe(onNext: { print("B ->", $0) })
print()

// Doesn't work as expected
// C -> (0, 1)
source
  .distinctUntilChanged { old, new in old == new }
  .subscribe(onNext: { print("C ->", $0) })
print()

// Works as expected
// D -> (0, 1)
// D -> (2, 3)
// D -> (4, 5)
source
  .distinctUntilChanged { "\($0)" == "\($1)" }
  .subscribe(onNext: { print("D ->", $0) })
print()

// Works as expected
// E -> (0, 1)
// E -> (2, 3)
// E -> (4, 5)
source
  .distinctUntilChanged { (); return $0 == $1 }
  .subscribe(onNext: { print("E ->", $0) })
print()

// Works as expected
// F -> (0, 1)
// F -> (2, 3)
// F -> (4, 5)
source
  .distinctUntilChanged { let _ = (); return $0 == $1 }
  .subscribe(onNext: { print("F ->", $0) })
print()

Related issue: ReactiveX/RxSwift#2097

How about

source
  .distinctUntilChanged { (old: (Int, Int), new: (Int, Int)) in old == new }
  .subscribe(onNext: { print("G ->", $0) })
print()

This case expected to following

// G -> (0, 1)
// G -> (2, 3)
// G -> (4, 5)

This case works as expected!
I tested this case with RxSwift 5.1.1 and Swift 5.1.3

Yeah that would definitely work, but what I wanted to do is to not specify types so that Swift compiler could infer the types.