If C++ fails when you do casts, why doesn't Haskell fail since it allows infinite recursion to build terms of whatever type you like? You can make the same argu

If C++ fails when you do casts, why doesn't Haskell fail since it allows infinite recursion to build terms of whatever type you like? You can make the same argument: "don't do that". C++ compilers can warn when you cast. Haskell compilers can warn on incomplete pattern matches. Is the definition you use not so broad as to admit all invariants being described as types? If a method requires a dictionary arguments with a certain key, is that a type to you? If you extend the term "type" to cover all invariants, I don't think that is the way that the term is commonly used, even though many invariants can be proven in e.g. dependently typed languages. All these terms are so wonky because a C++ type is not equal to a Haskell type. So I feel we can't ever have solid definitions of terms like "strong", since it depends what you compare it to, and it also depends what you compare from. So while X is strong in comparison to Y, that doesn't say much about X's relation to Z.