YES
* Step 1: TrivialSCCs YES
    + Considered Problem:
        Rules:
          0. evalterminatestart(A,B,C)    -> evalterminateentryin(A,B,C)        True                               (1,1)
          1. evalterminateentryin(A,B,C)  -> evalterminatebb1in(B,A,C)          True                               (?,1)
          2. evalterminatebb1in(A,B,C)    -> evalterminatebbin(A,B,C)           [100 >= B && A >= C]               (?,1)
          3. evalterminatebb1in(A,B,C)    -> evalterminatereturnin(A,B,C)       [B >= 101]                         (?,1)
          4. evalterminatebb1in(A,B,C)    -> evalterminatereturnin(A,B,C)       [C >= 1 + A]                       (?,1)
          5. evalterminatebbin(A,B,C)     -> evalterminatebb1in(-1 + A,C,1 + B) [A + -1*C >= 0 && 100 + -1*B >= 0] (?,1)
          6. evalterminatereturnin(A,B,C) -> evalterminatestop(A,B,C)           True                               (?,1)
        Signature:
          {(evalterminatebb1in,3)
          ;(evalterminatebbin,3)
          ;(evalterminateentryin,3)
          ;(evalterminatereturnin,3)
          ;(evalterminatestart,3)
          ;(evalterminatestop,3)}
        Flow Graph:
          [0->{1},1->{2,3,4},2->{5},3->{6},4->{6},5->{2,3,4},6->{}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: Looptree YES
    + Considered Problem:
        Rules:
          0. evalterminatestart(A,B,C)    -> evalterminateentryin(A,B,C)        True                               (1,1)
          1. evalterminateentryin(A,B,C)  -> evalterminatebb1in(B,A,C)          True                               (1,1)
          2. evalterminatebb1in(A,B,C)    -> evalterminatebbin(A,B,C)           [100 >= B && A >= C]               (?,1)
          3. evalterminatebb1in(A,B,C)    -> evalterminatereturnin(A,B,C)       [B >= 101]                         (1,1)
          4. evalterminatebb1in(A,B,C)    -> evalterminatereturnin(A,B,C)       [C >= 1 + A]                       (1,1)
          5. evalterminatebbin(A,B,C)     -> evalterminatebb1in(-1 + A,C,1 + B) [A + -1*C >= 0 && 100 + -1*B >= 0] (?,1)
          6. evalterminatereturnin(A,B,C) -> evalterminatestop(A,B,C)           True                               (1,1)
        Signature:
          {(evalterminatebb1in,3)
          ;(evalterminatebbin,3)
          ;(evalterminateentryin,3)
          ;(evalterminatereturnin,3)
          ;(evalterminatestart,3)
          ;(evalterminatestop,3)}
        Flow Graph:
          [0->{1},1->{2,3,4},2->{5},3->{6},4->{6},5->{2,3,4},6->{}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5,6]
          |
          `- p:[2,5] c: [5]

YES