I have created following item named BasicStyle.qml:

import QtQuick 2.0

Item {
	property color backgroundColor: "red"
}

and item OtherStyle.qml that 'derives' from it:

import QtQuick 2.0

BasicStyle {
	backgroundColor: "green"
}

Now I would like to use like this:

import QtQuick 2.12
import QtQuick.Window 2.12

Window {
	visible: true
	width: 640
	height: 480
	title: qsTr("Hello World")

//	property BasicStyle basicStyle: BasicStyle
	property BasicStyle basicStyle: OtherStyle

	color: basicStyle.backgroundColor
}

However all I can get is:

QML debugging is enabled. Only use this in a safe environment.
qrc:/main.qml:13: TypeError: Cannot read property 'backgroundColor' of null
qrc:/main.qml:11:2: Unable to assign QQuickWindowQmlImpl_QML_1 to BasicStyle_QMLTYPE_0

Do you have any idea what I'm missing here?