title: How erasure of generics "replace" multiple bounds
body: <p>I read that during the type erasure process, the Java compiler erases all type parameters and replaces each with its first bound if the type parameter is bounded, or Object if the type parameter is unbounded.
  But, i can't understand, <strong>isn't it redundant to specify the interface that the type parameter needs to implement</strong>. For example:</p>
  
  <pre><code>public class Box&lt;T extends Something,Seralizable,Cloneable&gt;
  </code></pre>
  
  <p>If erasure replaces <code>T</code> inside the class <code>Box</code> with <code>Something</code> (Class reference), doesn't it mean interfaces: <code>Seralizable,Cloneable</code> must be implemented by Something class, so only me it's feel redundant to specify <code>Seralizable,Cloneable</code> interfaces? In addition what happens if only interfaces are mentioned inside the diamonds, does by default <code>T</code> consider as <em><code>Object</code></em> reference?    </p>
  
  <p>I will be glad for one example for Generic Class and one example for Generic Method (if multiple extends exist in generics method).</p>