WORST_CASE(Omega(n^1),O(n^1))
* Step 1: Sum WORST_CASE(Omega(n^1),O(n^1))
    + Considered Problem:
        - Strict TRS:
            fold(a,xs) -> Cons(foldl(a,xs),Cons(foldr(a,xs),Nil()))
            foldl(a,Nil()) -> a
            foldl(x,Cons(S(0()),xs)) -> foldl(S(x),xs)
            foldl(S(0()),Cons(x,xs)) -> foldl(S(x),xs)
            foldr(a,Cons(x,xs)) -> op(x,foldr(a,xs))
            foldr(a,Nil()) -> a
            notEmpty(Cons(x,xs)) -> True()
            notEmpty(Nil()) -> False()
            op(x,S(0())) -> S(x)
            op(S(0()),y) -> S(y)
        - Signature:
            {fold/2,foldl/2,foldr/2,notEmpty/1,op/2} / {0/0,Cons/2,False/0,Nil/0,S/1,True/0}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {fold,foldl,foldr,notEmpty,op} and constructors {0,Cons
            ,False,Nil,S,True}
    + Applied Processor:
        Sum {left = someStrategy, right = someStrategy}
    + Details:
        ()
** Step 1.a:1: DecreasingLoops WORST_CASE(Omega(n^1),?)
    + Considered Problem:
        - Strict TRS:
            fold(a,xs) -> Cons(foldl(a,xs),Cons(foldr(a,xs),Nil()))
            foldl(a,Nil()) -> a
            foldl(x,Cons(S(0()),xs)) -> foldl(S(x),xs)
            foldl(S(0()),Cons(x,xs)) -> foldl(S(x),xs)
            foldr(a,Cons(x,xs)) -> op(x,foldr(a,xs))
            foldr(a,Nil()) -> a
            notEmpty(Cons(x,xs)) -> True()
            notEmpty(Nil()) -> False()
            op(x,S(0())) -> S(x)
            op(S(0()),y) -> S(y)
        - Signature:
            {fold/2,foldl/2,foldr/2,notEmpty/1,op/2} / {0/0,Cons/2,False/0,Nil/0,S/1,True/0}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {fold,foldl,foldr,notEmpty,op} and constructors {0,Cons
            ,False,Nil,S,True}
    + Applied Processor:
        DecreasingLoops {bound = AnyLoop, narrow = 10}
    + Details:
        The system has following decreasing Loops:
          foldl(x,y){y -> Cons(S(0()),y)} =
            foldl(x,Cons(S(0()),y)) ->^+ foldl(S(x),y)
              = C[foldl(S(x),y) = foldl(x,y){x -> S(x)}]

** Step 1.b:1: NaturalPI WORST_CASE(?,O(n^1))
    + Considered Problem:
        - Strict TRS:
            fold(a,xs) -> Cons(foldl(a,xs),Cons(foldr(a,xs),Nil()))
            foldl(a,Nil()) -> a
            foldl(x,Cons(S(0()),xs)) -> foldl(S(x),xs)
            foldl(S(0()),Cons(x,xs)) -> foldl(S(x),xs)
            foldr(a,Cons(x,xs)) -> op(x,foldr(a,xs))
            foldr(a,Nil()) -> a
            notEmpty(Cons(x,xs)) -> True()
            notEmpty(Nil()) -> False()
            op(x,S(0())) -> S(x)
            op(S(0()),y) -> S(y)
        - Signature:
            {fold/2,foldl/2,foldr/2,notEmpty/1,op/2} / {0/0,Cons/2,False/0,Nil/0,S/1,True/0}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {fold,foldl,foldr,notEmpty,op} and constructors {0,Cons
            ,False,Nil,S,True}
    + Applied Processor:
        NaturalPI {shape = Linear, restrict = Restrict, uargs = UArgs, urules = URules, selector = Just any strict-rules}
    + Details:
        We apply a polynomial interpretation of kind constructor-based(linear):
        The following argument positions are considered usable:
          uargs(Cons) = {1,2},
          uargs(op) = {2}
        
        Following symbols are considered usable:
          {fold,foldl,foldr,notEmpty,op}
        TcT has computed the following interpretation:
                 p(0) = 0               
              p(Cons) = 1 + x1 + x2     
             p(False) = 13              
               p(Nil) = 1               
                 p(S) = 0               
              p(True) = 10              
              p(fold) = 11 + 9*x1 + 7*x2
             p(foldl) = 4*x1 + 4*x2     
             p(foldr) = x1 + 3*x2       
          p(notEmpty) = 12 + 12*x1      
                p(op) = 2 + x2          
        
        Following rules are strictly oriented:
                      fold(a,xs) = 11 + 9*a + 7*xs                          
                                 > 3 + 5*a + 7*xs                           
                                 = Cons(foldl(a,xs),Cons(foldr(a,xs),Nil()))
        
                  foldl(a,Nil()) = 4 + 4*a                                  
                                 > a                                        
                                 = a                                        
        
        foldl(x,Cons(S(0()),xs)) = 4 + 4*x + 4*xs                           
                                 > 4*xs                                     
                                 = foldl(S(x),xs)                           
        
        foldl(S(0()),Cons(x,xs)) = 4 + 4*x + 4*xs                           
                                 > 4*xs                                     
                                 = foldl(S(x),xs)                           
        
             foldr(a,Cons(x,xs)) = 3 + a + 3*x + 3*xs                       
                                 > 2 + a + 3*xs                             
                                 = op(x,foldr(a,xs))                        
        
                  foldr(a,Nil()) = 3 + a                                    
                                 > a                                        
                                 = a                                        
        
            notEmpty(Cons(x,xs)) = 24 + 12*x + 12*xs                        
                                 > 10                                       
                                 = True()                                   
        
                 notEmpty(Nil()) = 24                                       
                                 > 13                                       
                                 = False()                                  
        
                    op(x,S(0())) = 2                                        
                                 > 0                                        
                                 = S(x)                                     
        
                    op(S(0()),y) = 2 + y                                    
                                 > 0                                        
                                 = S(y)                                     
        
        
        Following rules are (at-least) weakly oriented:
        
** Step 1.b:2: EmptyProcessor WORST_CASE(?,O(1))
    + Considered Problem:
        - Weak TRS:
            fold(a,xs) -> Cons(foldl(a,xs),Cons(foldr(a,xs),Nil()))
            foldl(a,Nil()) -> a
            foldl(x,Cons(S(0()),xs)) -> foldl(S(x),xs)
            foldl(S(0()),Cons(x,xs)) -> foldl(S(x),xs)
            foldr(a,Cons(x,xs)) -> op(x,foldr(a,xs))
            foldr(a,Nil()) -> a
            notEmpty(Cons(x,xs)) -> True()
            notEmpty(Nil()) -> False()
            op(x,S(0())) -> S(x)
            op(S(0()),y) -> S(y)
        - Signature:
            {fold/2,foldl/2,foldr/2,notEmpty/1,op/2} / {0/0,Cons/2,False/0,Nil/0,S/1,True/0}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {fold,foldl,foldr,notEmpty,op} and constructors {0,Cons
            ,False,Nil,S,True}
    + Applied Processor:
        EmptyProcessor
    + Details:
        The problem is already closed. The intended complexity is O(1).

WORST_CASE(Omega(n^1),O(n^1))