WORST_CASE(?,O(1))
* Step 1: RestrictVarsProcessor WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D) -> f8(0,B,C,D)         True                   (1,1)
          1. f8(A,B,C,D) -> f8(1 + A,B,C,D)     [3 >= A]               (?,1)
          2. f8(A,B,C,D) -> f8(1 + A,A,1 + A,E) [3 >= A]               (?,1)
          3. f8(A,B,C,D) -> f23(A,B,C,D)        [A >= 4 && 0 >= 1 + E] (?,1)
          4. f8(A,B,C,D) -> f23(A,B,C,D)        [A >= 4]               (?,1)
        Signature:
          {(f0,4);(f23,4);(f8,4)}
        Flow Graph:
          [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{},4->{}]
        
    + Applied Processor:
        RestrictVarsProcessor
    + Details:
        We removed the arguments [B,C,D] .
* Step 2: LocalSizeboundsProc WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A) -> f8(0)     True                   (1,1)
          1. f8(A) -> f8(1 + A) [3 >= A]               (?,1)
          2. f8(A) -> f8(1 + A) [3 >= A]               (?,1)
          3. f8(A) -> f23(A)    [A >= 4 && 0 >= 1 + E] (?,1)
          4. f8(A) -> f23(A)    [A >= 4]               (?,1)
        Signature:
          {(f0,1);(f23,1);(f8,1)}
        Flow Graph:
          [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{},4->{}]
        
    + Applied Processor:
        LocalSizeboundsProc
    + Details:
        LocalSizebounds generated; rvgraph
          (<0,0,A>,     0, .= 0) 
          (<1,0,A>, 1 + A, .+ 1) 
          (<2,0,A>, 1 + A, .+ 1) 
          (<3,0,A>,     A, .= 0) 
          (<4,0,A>,     A, .= 0) 
* Step 3: SizeboundsProc WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A) -> f8(0)     True                   (1,1)
          1. f8(A) -> f8(1 + A) [3 >= A]               (?,1)
          2. f8(A) -> f8(1 + A) [3 >= A]               (?,1)
          3. f8(A) -> f23(A)    [A >= 4 && 0 >= 1 + E] (?,1)
          4. f8(A) -> f23(A)    [A >= 4]               (?,1)
        Signature:
          {(f0,1);(f23,1);(f8,1)}
        Flow Graph:
          [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{},4->{}]
        Sizebounds:
          (<0,0,A>, ?) 
          (<1,0,A>, ?) 
          (<2,0,A>, ?) 
          (<3,0,A>, ?) 
          (<4,0,A>, ?) 
    + Applied Processor:
        SizeboundsProc
    + Details:
        Sizebounds computed:
          (<0,0,A>, 0) 
          (<1,0,A>, 4) 
          (<2,0,A>, 4) 
          (<3,0,A>, 4) 
          (<4,0,A>, 4) 
* Step 4: UnsatPaths WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A) -> f8(0)     True                   (1,1)
          1. f8(A) -> f8(1 + A) [3 >= A]               (?,1)
          2. f8(A) -> f8(1 + A) [3 >= A]               (?,1)
          3. f8(A) -> f23(A)    [A >= 4 && 0 >= 1 + E] (?,1)
          4. f8(A) -> f23(A)    [A >= 4]               (?,1)
        Signature:
          {(f0,1);(f23,1);(f8,1)}
        Flow Graph:
          [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{},4->{}]
        Sizebounds:
          (<0,0,A>, 0) 
          (<1,0,A>, 4) 
          (<2,0,A>, 4) 
          (<3,0,A>, 4) 
          (<4,0,A>, 4) 
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,3),(0,4)]
* Step 5: LeafRules WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A) -> f8(0)     True                   (1,1)
          1. f8(A) -> f8(1 + A) [3 >= A]               (?,1)
          2. f8(A) -> f8(1 + A) [3 >= A]               (?,1)
          3. f8(A) -> f23(A)    [A >= 4 && 0 >= 1 + E] (?,1)
          4. f8(A) -> f23(A)    [A >= 4]               (?,1)
        Signature:
          {(f0,1);(f23,1);(f8,1)}
        Flow Graph:
          [0->{1,2},1->{1,2,3,4},2->{1,2,3,4},3->{},4->{}]
        Sizebounds:
          (<0,0,A>, 0) 
          (<1,0,A>, 4) 
          (<2,0,A>, 4) 
          (<3,0,A>, 4) 
          (<4,0,A>, 4) 
    + Applied Processor:
        LeafRules
    + Details:
        The following transitions are estimated by its predecessors and are removed [3,4]
* Step 6: PolyRank WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A) -> f8(0)     True     (1,1)
          1. f8(A) -> f8(1 + A) [3 >= A] (?,1)
          2. f8(A) -> f8(1 + A) [3 >= A] (?,1)
        Signature:
          {(f0,1);(f23,1);(f8,1)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{1,2}]
        Sizebounds:
          (<0,0,A>, 0) 
          (<1,0,A>, 4) 
          (<2,0,A>, 4) 
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
          p(f0) = 4        
          p(f8) = 4 + -1*x1
        
        The following rules are strictly oriented:
        [3 >= A] ==>          
           f8(A)   = 4 + -1*A 
                   > 3 + -1*A 
                   = f8(1 + A)
        
        
        The following rules are weakly oriented:
            True ==>          
           f0(A)   = 4        
                  >= 4        
                   = f8(0)    
        
        [3 >= A] ==>          
           f8(A)   = 4 + -1*A 
                  >= 3 + -1*A 
                   = f8(1 + A)
        
        
* Step 7: PolyRank WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A) -> f8(0)     True     (1,1)
          1. f8(A) -> f8(1 + A) [3 >= A] (?,1)
          2. f8(A) -> f8(1 + A) [3 >= A] (4,1)
        Signature:
          {(f0,1);(f23,1);(f8,1)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{1,2}]
        Sizebounds:
          (<0,0,A>, 0) 
          (<1,0,A>, 4) 
          (<2,0,A>, 4) 
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
          p(f0) = 4        
          p(f8) = 4 + -1*x1
        
        The following rules are strictly oriented:
        [3 >= A] ==>          
           f8(A)   = 4 + -1*A 
                   > 3 + -1*A 
                   = f8(1 + A)
        
        [3 >= A] ==>          
           f8(A)   = 4 + -1*A 
                   > 3 + -1*A 
                   = f8(1 + A)
        
        
        The following rules are weakly oriented:
           True ==>      
          f0(A)   = 4    
                 >= 4    
                  = f8(0)
        
        
* Step 8: KnowledgePropagation WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A) -> f8(0)     True     (1,1)
          1. f8(A) -> f8(1 + A) [3 >= A] (4,1)
          2. f8(A) -> f8(1 + A) [3 >= A] (4,1)
        Signature:
          {(f0,1);(f23,1);(f8,1)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{1,2}]
        Sizebounds:
          (<0,0,A>, 0) 
          (<1,0,A>, 4) 
          (<2,0,A>, 4) 
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

WORST_CASE(?,O(1))