Deletion of the left-leaning Red Black tree and its invariant

Question

I have difficulty understanding the statement from this paper

To do so, we maintain the in-variant that the current node or its left child is red. We can do so by moving to the left unless the current node is red and its left child and left grandchild are both black. In that case, we can do a color flip, which restores the invariant but may introduce successive reds on the right

I understand that each path from a leaf node to the root must have the same number of black links/nodes, and that this paper's strategy is to color the one we want to delete with red and fix the tree thereafter. But I still have the following questions:

1. Why the invariant is the current node is red or the current.left is red?

2. more specifically, for the helper function moveRedLeft why it pays much attention to the color of current.left.right, if this node is red, we have to do 2 rotations and one flip.(the conditional statement for move move functions)

Answer 1

Why the invariant is the current node is red or the current.left is red?

This is a choice and characterizes the algorithm. There is of course no guaranty that this invariant is true if you just blindly move down the tree, as you risk to arrive at a spot where the current node and its left child are both black. But this algorithm sets as a goal to maintain this invariant, as it helps to stay within the rules of a valid black-red tree while still manipulating it.

Why can't we move to the left if the current = red, current.left = black, current.left.left = black?

If we would move to the left, then we would arrive in a state where both the current node and its left child would be black, which would violate the invariant.

Why a color flip would maintain this invariant?

When the algorithm arrives in the above "blocking" state, a color flip will turn the current red node black, and its children red. So the invariant is maintained (the left child is now red), and we can now move left while keeping the invariant: after moving to the left we get a state where the current node is red.

Note that this color flip might bring the tree in an invalid state, as (before moving down) the current.left.right node might be red, and as it is a child of a node that became red by the flip, this is now violating the rule that a red node may not have a red child. The article goes on to explain how the algorithm will resolve that invalid state...