Unsound subtyping behavior with (=> Any)

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The following is correctly disallowed:

scala> def foo(x: Int) = x + 1
foo: (Int)Int

scala> def bar(g: (=>Int) => Int, x: Int) = g(x)
bar: ((=> Int) => Int,Int)Int

scala> bar(foo, 3)
<console>:7: error: type mismatch;
 found   : (Int) => Int
 required: (=> Int) => Int
       bar(foo, 3)
           ^

but at present this seems to work:

scala> def myToString(x : Any) : String = x.toString   
foo: (Any)String

scala> def bar(g:(=>Any) => String, x:Int) = g(x)
bar: ((=> Any) => String,Int)String

scala> bar(myToString, 3)
res13: String = <function>

[scabug]

Imported From: https://issues.scala-lang.org/browse/SI-1386?orig=1
Reporter: Geoffrey Alan Washburn (washburn)
See #5517

[scabug]

@odersky said:
I fail to see the unsoundness. (=> Any) is a subtype of Any.

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Geoffrey Alan Washburn (washburn) said:
If (=> Any) is a subtype of Any, then why is (=> Int) not a subtype of Int?

Maybe you didn't look closely enough, but clearly the abstractions are getting violated because in the latter example myToString is receiving some kind of thunk rather than the integer 3 as its argument.

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Aaron Harnly (aaronharnly) said:
Replying to [comment:2 washburn]:

If (=> Any) is a subtype of Any, then why is (=> Int) not a subtype of Int?

Eh? (=> Any) is a subtype of Any in the same sense that (=> Int) is a subtype of Any.

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Geoffrey Alan Washburn (washburn) said:
Replying to [comment:3 aaronharnly]:

Replying to [comment:2 washburn]:

If (=> Any) is a subtype of Any, then why is (=> Int) not a subtype of Int?

Eh? (=> Any) is a subtype of Any in the same sense that (=> Int) is a subtype of Any.

Indeed, I hadn't been thinking about the problem in that fashion.

However, there is still the issue that the abstraction boundaries are getting violated. At least with respect the current description of by-name types/parameters in the specification.