YES
* Step 1: TrivialSCCs YES
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D,E) -> f1(F,1,0,D,E)            True                                     (1,1)
          1. f1(A,B,C,D,E) -> f1(-10 + A,-1 + B,C,D,E) [C >= 0 && B >= 1 && A >= 101]           (?,1)
          2. f1(A,B,C,D,E) -> f1(11 + A,1 + B,C,D,E)   [C >= 0 && B >= 1 && 100 >= A]           (?,1)
          3. f1(A,B,C,D,E) -> f1(-10 + A,-1 + B,1,A,B) [C >= 0 && A >= 101 && 0 >= C && B >= 1] (?,1)
          4. f1(A,B,C,D,E) -> f1(11 + A,1 + B,1,A,B)   [C >= 0 && 100 >= A && 0 >= C && B >= 1] (?,1)
          5. f1(A,B,C,D,E) -> f2(A,B,C,D,E)            [C >= 0 && D >= A && C >= 1 && B >= E]   (?,1)
        Signature:
          {(f0,5);(f1,5);(f2,5)}
        Flow Graph:
          [0->{1,2,3,4,5},1->{1,2,3,4,5},2->{1,2,3,4,5},3->{1,2,3,4,5},4->{1,2,3,4,5},5->{}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D,E) -> f1(F,1,0,D,E)            True                                     (1,1)
          1. f1(A,B,C,D,E) -> f1(-10 + A,-1 + B,C,D,E) [C >= 0 && B >= 1 && A >= 101]           (?,1)
          2. f1(A,B,C,D,E) -> f1(11 + A,1 + B,C,D,E)   [C >= 0 && B >= 1 && 100 >= A]           (?,1)
          3. f1(A,B,C,D,E) -> f1(-10 + A,-1 + B,1,A,B) [C >= 0 && A >= 101 && 0 >= C && B >= 1] (?,1)
          4. f1(A,B,C,D,E) -> f1(11 + A,1 + B,1,A,B)   [C >= 0 && 100 >= A && 0 >= C && B >= 1] (?,1)
          5. f1(A,B,C,D,E) -> f2(A,B,C,D,E)            [C >= 0 && D >= A && C >= 1 && B >= E]   (1,1)
        Signature:
          {(f0,5);(f1,5);(f2,5)}
        Flow Graph:
          [0->{1,2,3,4,5},1->{1,2,3,4,5},2->{1,2,3,4,5},3->{1,2,3,4,5},4->{1,2,3,4,5},5->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,5),(3,3),(3,4),(3,5),(4,3),(4,4),(4,5)]
* Step 3: Looptree YES
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D,E) -> f1(F,1,0,D,E)            True                                     (1,1)
          1. f1(A,B,C,D,E) -> f1(-10 + A,-1 + B,C,D,E) [C >= 0 && B >= 1 && A >= 101]           (?,1)
          2. f1(A,B,C,D,E) -> f1(11 + A,1 + B,C,D,E)   [C >= 0 && B >= 1 && 100 >= A]           (?,1)
          3. f1(A,B,C,D,E) -> f1(-10 + A,-1 + B,1,A,B) [C >= 0 && A >= 101 && 0 >= C && B >= 1] (?,1)
          4. f1(A,B,C,D,E) -> f1(11 + A,1 + B,1,A,B)   [C >= 0 && 100 >= A && 0 >= C && B >= 1] (?,1)
          5. f1(A,B,C,D,E) -> f2(A,B,C,D,E)            [C >= 0 && D >= A && C >= 1 && B >= E]   (1,1)
        Signature:
          {(f0,5);(f1,5);(f2,5)}
        Flow Graph:
          [0->{1,2,3,4},1->{1,2,3,4,5},2->{1,2,3,4,5},3->{1,2},4->{1,2},5->{}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5]
          |
          `- p:[1,2,3,4] c: [4]
             |
             `- p:[1,2,3] c: [3]
                |
                `- p:[1,2] c: [2]
                   |
                   `- p:[1] c: [1]

YES