scala - How to use a type constraint with abstract type -

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निम्न कोड को देखते हुए: 

  विशेषता एस {प्रकार टी} मामला वर्ग ए (टी : Seq [String]) S {type T = Seq [String]} का विस्तार करता है   
 मुझे इस संकलन की गलती नहीं है: ऐसा लगता है कि सबूत नहीं है।  <पूर्व>  def च [एस एंड एलटी:: ए, एक्स] (जी: = & gt; सीक [एक्स]) (अंतर्निहित ईव: एस # टी =: = सैक [एक्स]) = नया ए (जी) & lt; कंसोल>; 50: त्रुटि: प्रकार बेमेल; पाया: Seq [X] की आवश्यकता: Seq [स्ट्रिंग] डीईएफ़ च [एस & LT;: एक, X] (छ: = & gt; Seq [X]) (निहित EV: एस # टी =: = Seq [X]) नई = ए (जी)    
 
 
 यह कैसे मूर्ख लग सकता है,  =: =  ऑपेरेंड इस मुद्दे को हल करता है  
  डीफ़ एफ [एस एंड एलटी;: ए, एक्स] (जी: = & gt; सीईसी [एक्स]) (असल ev: SEQ [X ] =: = एस # टी) = नया ए (जी)   
 धन्यवाद, स्केलैक।  
 स्पष्टीकरण  
 जब आप कहते हैं  अंतर्निहित उत्प्रवास एस # टी =: = सैक [एक्स] , संकलक  S # T  से  Seq [X]  से एक अंतर्निहित रूपांतरण प्रदान करता है। हालांकि यह  सेक् [एक्स]  से  एस # टी  के बराबर रूपांतरण प्रदान नहीं करता है, और यह मूर्खतापूर्ण हिस्सा है: समानता को कम्यूटेट करने वाला नहीं है?  
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