TypeConstructor's equality for classes relies on FqNames

This is needed because different modules/libraries may define classes with the same FqNames, which may be identical or slightly different.
 Such classes must be considered equal, because your dependencies may rely on different packagings of the same codebase, and the classes there will be distinct though identical
  (think intellij-core vs idea-full).

core/descriptor.loader.java/src/org/jetbrains/jet/lang/resolve/java/lazy/descriptors/LazyJavaClassDescriptor.kt

@@ -52,6 +52,7 @@ import org.jetbrains.jet.lang.resolve.java.resolver.DescriptorResolverUtils
 import org.jetbrains.jet.lang.resolve.java.descriptor.JavaClassStaticsPackageFragmentDescriptor
 import org.jetbrains.jet.lang.descriptors.impl.MutableClassDescriptor
 import org.jetbrains.jet.lang.resolve.name.SpecialNames
+import org.jetbrains.jet.lang.types.AbstractClassTypeConstructor
 
 class LazyJavaClassDescriptor(
         private val outerC: LazyJavaResolverContextWithTypes,
@@ -188,7 +189,7 @@ class LazyJavaClassDescriptor(
 
     override fun toString() = "lazy java class $fqName"
 
-    private inner class LazyJavaClassTypeConstructor : TypeConstructor {
+    private inner class LazyJavaClassTypeConstructor : AbstractClassTypeConstructor() {
 
         private val _parameters = c.storageManager.createLazyValue {
             jClass.getTypeParameters().map({