MAYBE
* Step 1: UnsatPaths MAYBE
    + Considered Problem:
        Rules:
          0. f300(A,B,C) -> f2(A,B,C)      True                   (1,1)
          1. f2(A,B,C)   -> f1(A,0,D)      [0 >= A]               (?,1)
          2. f2(A,B,C)   -> f2(A,0,C)      [A >= 1]               (?,1)
          3. f2(A,B,C)   -> f1(-1 + A,D,E) [0 >= 1 + D && 1 >= A] (?,1)
          4. f2(A,B,C)   -> f1(-1 + A,D,E) [D >= 1 && 1 >= A]     (?,1)
          5. f2(A,B,C)   -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 2] (?,1)
          6. f2(A,B,C)   -> f2(-1 + A,D,C) [D >= 1 && A >= 2]     (?,1)
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Flow Graph:
          [0->{1,2,3,4,5,6},1->{},2->{1,2,3,4,5,6},3->{},4->{},5->{1,2,3,4,5,6},6->{1,2,3,4,5,6}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(2,1),(5,1),(6,1)]
* Step 2: TrivialSCCs MAYBE
    + Considered Problem:
        Rules:
          0. f300(A,B,C) -> f2(A,B,C)      True                   (1,1)
          1. f2(A,B,C)   -> f1(A,0,D)      [0 >= A]               (?,1)
          2. f2(A,B,C)   -> f2(A,0,C)      [A >= 1]               (?,1)
          3. f2(A,B,C)   -> f1(-1 + A,D,E) [0 >= 1 + D && 1 >= A] (?,1)
          4. f2(A,B,C)   -> f1(-1 + A,D,E) [D >= 1 && 1 >= A]     (?,1)
          5. f2(A,B,C)   -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 2] (?,1)
          6. f2(A,B,C)   -> f2(-1 + A,D,C) [D >= 1 && A >= 2]     (?,1)
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Flow Graph:
          [0->{1,2,3,4,5,6},1->{},2->{2,3,4,5,6},3->{},4->{},5->{2,3,4,5,6},6->{2,3,4,5,6}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 3: PolyRank MAYBE
    + Considered Problem:
        Rules:
          0. f300(A,B,C) -> f2(A,B,C)      True                   (1,1)
          1. f2(A,B,C)   -> f1(A,0,D)      [0 >= A]               (1,1)
          2. f2(A,B,C)   -> f2(A,0,C)      [A >= 1]               (?,1)
          3. f2(A,B,C)   -> f1(-1 + A,D,E) [0 >= 1 + D && 1 >= A] (1,1)
          4. f2(A,B,C)   -> f1(-1 + A,D,E) [D >= 1 && 1 >= A]     (1,1)
          5. f2(A,B,C)   -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 2] (?,1)
          6. f2(A,B,C)   -> f2(-1 + A,D,C) [D >= 1 && A >= 2]     (?,1)
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Flow Graph:
          [0->{1,2,3,4,5,6},1->{},2->{2,3,4,5,6},3->{},4->{},5->{2,3,4,5,6},6->{2,3,4,5,6}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
            p(f1) = x1
            p(f2) = x1
          p(f300) = x1
        
        Following rules are strictly oriented:
        [D >= 1 && A >= 2] ==>               
                 f2(A,B,C)   = A             
                             > -1 + A        
                             = f2(-1 + A,D,C)
        
        
        Following rules are weakly oriented:
                          True ==>               
                   f300(A,B,C)   = A             
                                >= A             
                                 = f2(A,B,C)     
        
                      [0 >= A] ==>               
                     f2(A,B,C)   = A             
                                >= A             
                                 = f1(A,0,D)     
        
                      [A >= 1] ==>               
                     f2(A,B,C)   = A             
                                >= A             
                                 = f2(A,0,C)     
        
        [0 >= 1 + D && 1 >= A] ==>               
                     f2(A,B,C)   = A             
                                >= -1 + A        
                                 = f1(-1 + A,D,E)
        
            [D >= 1 && 1 >= A] ==>               
                     f2(A,B,C)   = A             
                                >= -1 + A        
                                 = f1(-1 + A,D,E)
        
        [0 >= 1 + D && A >= 2] ==>               
                     f2(A,B,C)   = A             
                                >= -1 + A        
                                 = f2(-1 + A,D,C)
        
        
* Step 4: PolyRank MAYBE
    + Considered Problem:
        Rules:
          0. f300(A,B,C) -> f2(A,B,C)      True                   (1,1)
          1. f2(A,B,C)   -> f1(A,0,D)      [0 >= A]               (1,1)
          2. f2(A,B,C)   -> f2(A,0,C)      [A >= 1]               (?,1)
          3. f2(A,B,C)   -> f1(-1 + A,D,E) [0 >= 1 + D && 1 >= A] (1,1)
          4. f2(A,B,C)   -> f1(-1 + A,D,E) [D >= 1 && 1 >= A]     (1,1)
          5. f2(A,B,C)   -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 2] (?,1)
          6. f2(A,B,C)   -> f2(-1 + A,D,C) [D >= 1 && A >= 2]     (A,1)
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Flow Graph:
          [0->{1,2,3,4,5,6},1->{},2->{2,3,4,5,6},3->{},4->{},5->{2,3,4,5,6},6->{2,3,4,5,6}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
            p(f1) = x1
            p(f2) = x1
          p(f300) = x1
        
        Following rules are strictly oriented:
        [0 >= 1 + D && A >= 2] ==>               
                     f2(A,B,C)   = A             
                                 > -1 + A        
                                 = f2(-1 + A,D,C)
        
            [D >= 1 && A >= 2] ==>               
                     f2(A,B,C)   = A             
                                 > -1 + A        
                                 = f2(-1 + A,D,C)
        
        
        Following rules are weakly oriented:
                          True ==>               
                   f300(A,B,C)   = A             
                                >= A             
                                 = f2(A,B,C)     
        
                      [0 >= A] ==>               
                     f2(A,B,C)   = A             
                                >= A             
                                 = f1(A,0,D)     
        
                      [A >= 1] ==>               
                     f2(A,B,C)   = A             
                                >= A             
                                 = f2(A,0,C)     
        
        [0 >= 1 + D && 1 >= A] ==>               
                     f2(A,B,C)   = A             
                                >= -1 + A        
                                 = f1(-1 + A,D,E)
        
            [D >= 1 && 1 >= A] ==>               
                     f2(A,B,C)   = A             
                                >= -1 + A        
                                 = f1(-1 + A,D,E)
        
        
* Step 5: Failure MAYBE
  + Considered Problem:
      Rules:
        0. f300(A,B,C) -> f2(A,B,C)      True                   (1,1)
        1. f2(A,B,C)   -> f1(A,0,D)      [0 >= A]               (1,1)
        2. f2(A,B,C)   -> f2(A,0,C)      [A >= 1]               (?,1)
        3. f2(A,B,C)   -> f1(-1 + A,D,E) [0 >= 1 + D && 1 >= A] (1,1)
        4. f2(A,B,C)   -> f1(-1 + A,D,E) [D >= 1 && 1 >= A]     (1,1)
        5. f2(A,B,C)   -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 2] (A,1)
        6. f2(A,B,C)   -> f2(-1 + A,D,C) [D >= 1 && A >= 2]     (A,1)
      Signature:
        {(f1,3);(f2,3);(f300,3)}
      Flow Graph:
        [0->{1,2,3,4,5,6},1->{},2->{2,3,4,5,6},3->{},4->{},5->{2,3,4,5,6},6->{2,3,4,5,6}]
      
  + Applied Processor:
      Failing "Open problems left."
  + Details:
      Open problems left.
MAYBE