YES
* Step 1: Looptree YES
    + Considered Problem:
        Rules:
          0. eval1(A,B) -> eval2(A,1)      [A >= 0]                                                           (?,1)
          1. eval2(A,B) -> eval2(A,2*B)    [-1 + B >= 0 && -1 + A + B >= 0 && A >= 0 && B >= 1 && A >= 1 + B] (?,1)
          2. eval2(A,B) -> eval1(-1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && A >= 0 && B >= 1 && B >= A]     (?,1)
          3. start(A,B) -> eval1(A,B)      True                                                               (1,1)
        Signature:
          {(eval1,2);(eval2,2);(start,2)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{0},3->{0}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3]
          |
          `- p:[0,2,1] c: [2]
             |
             `- p:[1] c: [1]

YES