YES
* Step 1: TrivialSCCs YES
    + Considered Problem:
        Rules:
          0.  evalSequentialSinglestart(A,B)    -> evalSequentialSingleentryin(A,B)   True                                                                           (1,1)
          1.  evalSequentialSingleentryin(A,B)  -> evalSequentialSinglebb1in(0,B)     True                                                                           (?,1)
          2.  evalSequentialSinglebb1in(A,B)    -> evalSequentialSinglebb5in(A,B)     [A >= 0 && A >= B]                                                             (?,1)
          3.  evalSequentialSinglebb1in(A,B)    -> evalSequentialSinglebb2in(A,B)     [A >= 0 && B >= 1 + A]                                                         (?,1)
          4.  evalSequentialSinglebb2in(A,B)    -> evalSequentialSinglebbin(A,B)      [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + C] (?,1)
          5.  evalSequentialSinglebb2in(A,B)    -> evalSequentialSinglebbin(A,B)      [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1]     (?,1)
          6.  evalSequentialSinglebb2in(A,B)    -> evalSequentialSinglebb5in(A,B)     [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0]               (?,1)
          7.  evalSequentialSinglebbin(A,B)     -> evalSequentialSinglebb1in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0]               (?,1)
          8.  evalSequentialSinglebb5in(A,B)    -> evalSequentialSinglebb4in(A,B)     [A >= 0 && B >= 1 + A]                                                         (?,1)
          9.  evalSequentialSinglebb5in(A,B)    -> evalSequentialSinglereturnin(A,B)  [A >= 0 && A >= B]                                                             (?,1)
          10. evalSequentialSinglebb4in(A,B)    -> evalSequentialSinglebb5in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0]               (?,1)
          11. evalSequentialSinglereturnin(A,B) -> evalSequentialSinglestop(A,B)      [A + -1*B >= 0 && A >= 0]                                                      (?,1)
        Signature:
          {(evalSequentialSinglebb1in,2)
          ;(evalSequentialSinglebb2in,2)
          ;(evalSequentialSinglebb4in,2)
          ;(evalSequentialSinglebb5in,2)
          ;(evalSequentialSinglebbin,2)
          ;(evalSequentialSingleentryin,2)
          ;(evalSequentialSinglereturnin,2)
          ;(evalSequentialSinglestart,2)
          ;(evalSequentialSinglestop,2)}
        Flow Graph:
          [0->{1},1->{2,3},2->{8,9},3->{4,5,6},4->{7},5->{7},6->{8,9},7->{2,3},8->{10},9->{11},10->{8,9},11->{}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: UnsatPaths YES
    + Considered Problem:
        Rules:
          0.  evalSequentialSinglestart(A,B)    -> evalSequentialSingleentryin(A,B)   True                                                                           (1,1)
          1.  evalSequentialSingleentryin(A,B)  -> evalSequentialSinglebb1in(0,B)     True                                                                           (1,1)
          2.  evalSequentialSinglebb1in(A,B)    -> evalSequentialSinglebb5in(A,B)     [A >= 0 && A >= B]                                                             (1,1)
          3.  evalSequentialSinglebb1in(A,B)    -> evalSequentialSinglebb2in(A,B)     [A >= 0 && B >= 1 + A]                                                         (?,1)
          4.  evalSequentialSinglebb2in(A,B)    -> evalSequentialSinglebbin(A,B)      [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + C] (?,1)
          5.  evalSequentialSinglebb2in(A,B)    -> evalSequentialSinglebbin(A,B)      [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1]     (?,1)
          6.  evalSequentialSinglebb2in(A,B)    -> evalSequentialSinglebb5in(A,B)     [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0]               (1,1)
          7.  evalSequentialSinglebbin(A,B)     -> evalSequentialSinglebb1in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0]               (?,1)
          8.  evalSequentialSinglebb5in(A,B)    -> evalSequentialSinglebb4in(A,B)     [A >= 0 && B >= 1 + A]                                                         (?,1)
          9.  evalSequentialSinglebb5in(A,B)    -> evalSequentialSinglereturnin(A,B)  [A >= 0 && A >= B]                                                             (1,1)
          10. evalSequentialSinglebb4in(A,B)    -> evalSequentialSinglebb5in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0]               (?,1)
          11. evalSequentialSinglereturnin(A,B) -> evalSequentialSinglestop(A,B)      [A + -1*B >= 0 && A >= 0]                                                      (1,1)
        Signature:
          {(evalSequentialSinglebb1in,2)
          ;(evalSequentialSinglebb2in,2)
          ;(evalSequentialSinglebb4in,2)
          ;(evalSequentialSinglebb5in,2)
          ;(evalSequentialSinglebbin,2)
          ;(evalSequentialSingleentryin,2)
          ;(evalSequentialSinglereturnin,2)
          ;(evalSequentialSinglestart,2)
          ;(evalSequentialSinglestop,2)}
        Flow Graph:
          [0->{1},1->{2,3},2->{8,9},3->{4,5,6},4->{7},5->{7},6->{8,9},7->{2,3},8->{10},9->{11},10->{8,9},11->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(2,8),(6,9)]
* Step 3: Looptree YES
    + Considered Problem:
        Rules:
          0.  evalSequentialSinglestart(A,B)    -> evalSequentialSingleentryin(A,B)   True                                                                           (1,1)
          1.  evalSequentialSingleentryin(A,B)  -> evalSequentialSinglebb1in(0,B)     True                                                                           (1,1)
          2.  evalSequentialSinglebb1in(A,B)    -> evalSequentialSinglebb5in(A,B)     [A >= 0 && A >= B]                                                             (1,1)
          3.  evalSequentialSinglebb1in(A,B)    -> evalSequentialSinglebb2in(A,B)     [A >= 0 && B >= 1 + A]                                                         (?,1)
          4.  evalSequentialSinglebb2in(A,B)    -> evalSequentialSinglebbin(A,B)      [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + C] (?,1)
          5.  evalSequentialSinglebb2in(A,B)    -> evalSequentialSinglebbin(A,B)      [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1]     (?,1)
          6.  evalSequentialSinglebb2in(A,B)    -> evalSequentialSinglebb5in(A,B)     [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0]               (1,1)
          7.  evalSequentialSinglebbin(A,B)     -> evalSequentialSinglebb1in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0]               (?,1)
          8.  evalSequentialSinglebb5in(A,B)    -> evalSequentialSinglebb4in(A,B)     [A >= 0 && B >= 1 + A]                                                         (?,1)
          9.  evalSequentialSinglebb5in(A,B)    -> evalSequentialSinglereturnin(A,B)  [A >= 0 && A >= B]                                                             (1,1)
          10. evalSequentialSinglebb4in(A,B)    -> evalSequentialSinglebb5in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0]               (?,1)
          11. evalSequentialSinglereturnin(A,B) -> evalSequentialSinglestop(A,B)      [A + -1*B >= 0 && A >= 0]                                                      (1,1)
        Signature:
          {(evalSequentialSinglebb1in,2)
          ;(evalSequentialSinglebb2in,2)
          ;(evalSequentialSinglebb4in,2)
          ;(evalSequentialSinglebb5in,2)
          ;(evalSequentialSinglebbin,2)
          ;(evalSequentialSingleentryin,2)
          ;(evalSequentialSinglereturnin,2)
          ;(evalSequentialSinglestart,2)
          ;(evalSequentialSinglestop,2)}
        Flow Graph:
          [0->{1},1->{2,3},2->{9},3->{4,5,6},4->{7},5->{7},6->{8},7->{2,3},8->{10},9->{11},10->{8,9},11->{}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8,9,10,11]
          |
          +- p:[3,7,4,5] c: [7]
          |
          `- p:[8,10] c: [10]

YES