YES
* Step 1: TrivialSCCs YES
    + Considered Problem:
        Rules:
          0. f2(A,B,C,D,E)   -> f1(A,F,C,D,E)           [1 >= A]                                                                    (1,1)
          1. f2(A,B,C,D,E)   -> f300(A,B,C,D,E)         [A >= 2 && C >= 2]                                                          (1,1)
          2. f300(A,B,C,D,E) -> f1(A,F,C,D,E)           [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0]               (?,1)
          3. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D]     (?,1)
          4. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] (?,1)
        Signature:
          {(f1,5);(f2,5);(f300,5)}
        Flow Graph:
          [0->{},1->{2,3,4},2->{},3->{2,3,4},4->{2,3,4}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. f2(A,B,C,D,E)   -> f1(A,F,C,D,E)           [1 >= A]                                                                    (1,1)
          1. f2(A,B,C,D,E)   -> f300(A,B,C,D,E)         [A >= 2 && C >= 2]                                                          (1,1)
          2. f300(A,B,C,D,E) -> f1(A,F,C,D,E)           [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0]               (1,1)
          3. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D]     (?,1)
          4. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] (?,1)
        Signature:
          {(f1,5);(f2,5);(f300,5)}
        Flow Graph:
          [0->{},1->{2,3,4},2->{},3->{2,3,4},4->{2,3,4}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(3,4),(4,3)]
* Step 3: Looptree YES
    + Considered Problem:
        Rules:
          0. f2(A,B,C,D,E)   -> f1(A,F,C,D,E)           [1 >= A]                                                                    (1,1)
          1. f2(A,B,C,D,E)   -> f300(A,B,C,D,E)         [A >= 2 && C >= 2]                                                          (1,1)
          2. f300(A,B,C,D,E) -> f1(A,F,C,D,E)           [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0]               (1,1)
          3. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D]     (?,1)
          4. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] (?,1)
        Signature:
          {(f1,5);(f2,5);(f300,5)}
        Flow Graph:
          [0->{},1->{2,3,4},2->{},3->{2,3},4->{2,4}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4]
          |
          +- p:[4] c: [4]
          |
          `- p:[3] c: [3]

YES