How remove_reference disable template argument deductions?

According to this link, template argument deduction disallowed for std::forward and std::remove_reference is helping us to achieve that. But how does using remove_reference prevent template deduction from happening here?

template <class S>
S&& forward(typename std::remove_reference<S>::type& t) noexcept
{
    return static_cast<S&&>(t);
}