implicit convert from lambda to std::function inside template

I want to implement a template function which takes a lambda as argument.

#include <functional>

template<typename ... Result> using Fun = std::function<void(Result ...)>;
template<typename ... Result> void yield(Fun<Result ...>&& body) {};

template <typename T>
struct identity {
    typedef T type;
};

template<typename ... Result> void yield2(typename identity<Fun<Result ...>>::type && body) {};

int main() {
    yield<char>(
        Fun<char>(
            [](char) -> void {} // 1. success
        )
    );

    yield2<char>(
        [](char) -> void {} // 2. success with identify
    );

    yield<char>(
        [](char) -> void {} // 3. fail, seems achievable
    );

    yield(
        [](char) -> void {} // 4. fail, impossible ?
    );

    return 0;
}

Why case 3 fails ? I already giving the template parameters to the template, so it should be able to deduce the function type and implicit convert the lambda to function

EDIT:

Compiler always reduce template parameters from function arguments, can we reverse it ?

    yield<char>(
        [](auto&& c) -> void {} // 5. is it possible ?
    );