YES
* Step 1: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. eval(A,B)  -> eval(A,A)      [0 >= A && B = 1]                    (?,1)
          1. eval(A,B)  -> eval(A,A)      [B >= 1 && 1 + B >= 0 && B >= 1 + A] (?,1)
          2. eval(A,B)  -> eval(A,0)      [A >= 1 && B = 1]                    (?,1)
          3. eval(A,B)  -> eval(A,-1 + B) [B >= 1 && 1 + B >= 0 && A >= B]     (?,1)
          4. start(A,B) -> eval(A,B)      True                                 (1,1)
        Signature:
          {(eval,2);(start,2)}
        Flow Graph:
          [0->{0,1,2,3},1->{0,1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,0)
                                                             ,(0,1)
                                                             ,(0,2)
                                                             ,(0,3)
                                                             ,(1,0)
                                                             ,(1,1)
                                                             ,(2,0)
                                                             ,(2,1)
                                                             ,(2,2)
                                                             ,(2,3)
                                                             ,(3,0)
                                                             ,(3,1)]
* Step 2: FromIts YES
    + Considered Problem:
        Rules:
          0. eval(A,B)  -> eval(A,A)      [0 >= A && B = 1]                    (?,1)
          1. eval(A,B)  -> eval(A,A)      [B >= 1 && 1 + B >= 0 && B >= 1 + A] (?,1)
          2. eval(A,B)  -> eval(A,0)      [A >= 1 && B = 1]                    (?,1)
          3. eval(A,B)  -> eval(A,-1 + B) [B >= 1 && 1 + B >= 0 && A >= B]     (?,1)
          4. start(A,B) -> eval(A,B)      True                                 (1,1)
        Signature:
          {(eval,2);(start,2)}
        Flow Graph:
          [0->{},1->{2,3},2->{},3->{2,3},4->{0,1,2,3}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 3: Decompose YES
    + Considered Problem:
        Rules:
          eval(A,B)  -> eval(A,A)       [0 >= A && B = 1]
          eval(A,B)  -> eval(A,A)       [B >= 1 && 1 + B >= 0 && B >= 1 + A]
          eval(A,B)  -> eval(A,0)       [A >= 1 && B = 1]
          eval(A,B)  -> eval(A,-1 + B)  [B >= 1 && 1 + B >= 0 && A >= B]
          start(A,B) -> eval(A,B)       True
        Signature:
          {(eval,2);(start,2)}
        Rule Graph:
          [0->{},1->{2,3},2->{},3->{2,3},4->{0,1,2,3}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4]
          |
          `- p:[3] c: [3]
* Step 4: CloseWith YES
    + Considered Problem:
        (Rules:
          eval(A,B)  -> eval(A,A)       [0 >= A && B = 1]
          eval(A,B)  -> eval(A,A)       [B >= 1 && 1 + B >= 0 && B >= 1 + A]
          eval(A,B)  -> eval(A,0)       [A >= 1 && B = 1]
          eval(A,B)  -> eval(A,-1 + B)  [B >= 1 && 1 + B >= 0 && A >= B]
          start(A,B) -> eval(A,B)       True
        Signature:
          {(eval,2);(start,2)}
        Rule Graph:
          [0->{},1->{2,3},2->{},3->{2,3},4->{0,1,2,3}]
        ,We construct a looptree:
          P: [0,1,2,3,4]
          |
          `- p:[3] c: [3])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES