I found that the following works:
While 1 Menu("Test","A",A,"B",B,"C",C) Lbl A A+1→ A End Lbl B B+1→ B End Lbl C C+1→ C End
No matter what choice you choose, the End is always interpreted as the end of the while loop and it always loop back, it doesn't matter that they're different Ends. Therefore, you can have more than one End for every loop, and even possibly more than one loop for every End. This could have some profound impact on optimization, speed, and loop/label coding.