Matching alias template as template argument

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This is CWG issue 1286. The question is: are alias and test equivalent? There used to be an example in [temp.type] which suggested that y and z have the same type here:

template<template<class> class TT> struct X { };
template<class> struct Y { };
template<class T> using Z = Y<T>;
X<Y> y;
X<Z> z;

The example was corrected as part of CWG defect 1244 – which indicated correctly that there is no wording in [temp.alias] that actually specifies that alias templates are equivalent to the templates they alias. The only wording there refers to equivalence of alias template specializations:

When a template-id refers to the specialization of an alias template, it is equivalent to the associated type obtained by substitution of its template-arguments for the template-parameters in the type-id of the alias template.

The intent is apparently that y and z do have the same type in this example, meaning that Z and Y are actually equivalent. But unless and until the wording in the resolution is adopted, they are not. Today, alias and test are not equivalent but alias<int> and test<int> are. This means that is_specialization_of<alias, alias<int>> is is_specialization_of<alias, test<int>>, where alias is unique from test, which would not match your partial specialization and thus be false_type.

Moreover, even with the adoption of the wording in #1286, test and alias are still not equivalent for the obvious reason that test takes two template parameters and alias takes one template parameter. The example in the resolution wording mimics your example and clarifies the intent here:

template<typename T, U = T> struct A;

// ...

template<typename V>   
  using D = A<V>;      // not equivalent to A:
                       // different number of parameters

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