Proving correctness for greedy algorithm in string removal problem

Problem Statement: You are given a string s and two integers x and y. You can perform two types of operations any number of times. Remove substring "ab" and gain x points. For example, when removing "ab" from "cabxbae" it becomes "cxbae". Remove substring "ba" and gain y points. For example, when removing "ba" from "cabxbae" it becomes "cabxe". Return the maximum points you can gain after applying the above operations on s.

A greedy algorithm solves this problem in O(n). The greedy algorithm works by eliminating all pairs with a higher scoring value before removing pairs of a lower scoring value. Can anyone present a proof for the correctness of a greedy algorithm here. I have been trying, and I cannot convince myself.

(For the greedy algorithm described to be optimal, it is also necessary to assume that $$x$$ and $$y$$ are nonnegative.)
Without loss of generality, we can assume $$y \ge x$$.
Suppose to the contrary that the greedy algorithm is not optimal. Then there exist some string (possibly an intermediate state) containing a consecutive pair $$\mathrm{ba}$$ whose removal is suboptimal. Consider what happens to those letters in the optimal solution.
• If exactly one of those letters is removed in the optimal solution, then modifying the solution by first removing the $$\mathrm{ba}$$ pair, and omitting whatever removal applied to one of them later, gives another solution that is at least as good, contradicting the assumed suboptimality of removing the $$\mathrm{ba}$$ pair. (Because the other member of the pair was never removed in the optimal solution, this change works, in that it cannot break any later removals.)
• If both of those letters are removed in the optimal solution, that means they are paired up with an $$\mathrm{a}$$ on the left and a $$\mathrm{b}$$ on the right. In this case, we can modify the solution by first removing the $$\mathrm{ba}$$ pair, and later pairing up and removing those other $$\mathrm{a}$$ and $$\mathrm{b}$$, which is again a contradiction.