Remove spurious type check in is_l1_renamed

There is no reason for an expression renamed to level 1 to have type
renamed to level 2.

Author: romainbrenguier

src/goto-symex/renamed.cpp

@@ -15,9 +15,6 @@ Author: Romain Brenguier, [email protected]
 
 bool is_l1_renamed(const exprt &expr)
 {
-  if(!is_l2_renamed(expr.type()))
-    return false;
-
   if(expr.id() == ID_symbol)
   {
     const auto &type = expr.type();