WORST_CASE(Omega(n^1),O(n^1))
* Step 1: Sum WORST_CASE(Omega(n^1),O(n^1))
    + Considered Problem:
        - Strict TRS:
            eq0(0(),0()) -> S(0())
            eq0(0(),S(x)) -> 0()
            eq0(S(x),0()) -> 0()
            eq0(S(x'),S(x)) -> eq0(x',x)
        - Signature:
            {eq0/2} / {0/0,S/1}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {eq0} and constructors {0,S}
    + Applied Processor:
        Sum {left = someStrategy, right = someStrategy}
    + Details:
        ()
** Step 1.a:1: DecreasingLoops WORST_CASE(Omega(n^1),?)
    + Considered Problem:
        - Strict TRS:
            eq0(0(),0()) -> S(0())
            eq0(0(),S(x)) -> 0()
            eq0(S(x),0()) -> 0()
            eq0(S(x'),S(x)) -> eq0(x',x)
        - Signature:
            {eq0/2} / {0/0,S/1}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {eq0} and constructors {0,S}
    + Applied Processor:
        DecreasingLoops {bound = AnyLoop, narrow = 10}
    + Details:
        The system has following decreasing Loops:
          eq0(x,y){x -> S(x),y -> S(y)} =
            eq0(S(x),S(y)) ->^+ eq0(x,y)
              = C[eq0(x,y) = eq0(x,y){}]

** Step 1.b:1: NaturalPI WORST_CASE(?,O(n^1))
    + Considered Problem:
        - Strict TRS:
            eq0(0(),0()) -> S(0())
            eq0(0(),S(x)) -> 0()
            eq0(S(x),0()) -> 0()
            eq0(S(x'),S(x)) -> eq0(x',x)
        - Signature:
            {eq0/2} / {0/0,S/1}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {eq0} and constructors {0,S}
    + 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:
          none
        
        Following symbols are considered usable:
          {eq0}
        TcT has computed the following interpretation:
            p(0) = 2
            p(S) = 0
          p(eq0) = 9
        
        Following rules are strictly oriented:
         eq0(0(),0()) = 9     
                      > 0     
                      = S(0())
        
        eq0(0(),S(x)) = 9     
                      > 2     
                      = 0()   
        
        eq0(S(x),0()) = 9     
                      > 2     
                      = 0()   
        
        
        Following rules are (at-least) weakly oriented:
        eq0(S(x'),S(x)) =  9        
                        >= 9        
                        =  eq0(x',x)
        
** Step 1.b:2: NaturalPI WORST_CASE(?,O(n^1))
    + Considered Problem:
        - Strict TRS:
            eq0(S(x'),S(x)) -> eq0(x',x)
        - Weak TRS:
            eq0(0(),0()) -> S(0())
            eq0(0(),S(x)) -> 0()
            eq0(S(x),0()) -> 0()
        - Signature:
            {eq0/2} / {0/0,S/1}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {eq0} and constructors {0,S}
    + 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:
          none
        
        Following symbols are considered usable:
          {eq0}
        TcT has computed the following interpretation:
            p(0) = 3     
            p(S) = 2 + x1
          p(eq0) = 9*x2  
        
        Following rules are strictly oriented:
        eq0(S(x'),S(x)) = 18 + 9*x 
                        > 9*x      
                        = eq0(x',x)
        
        
        Following rules are (at-least) weakly oriented:
         eq0(0(),0()) =  27      
                      >= 5       
                      =  S(0())  
        
        eq0(0(),S(x)) =  18 + 9*x
                      >= 3       
                      =  0()     
        
        eq0(S(x),0()) =  27      
                      >= 3       
                      =  0()     
        
** Step 1.b:3: EmptyProcessor WORST_CASE(?,O(1))
    + Considered Problem:
        - Weak TRS:
            eq0(0(),0()) -> S(0())
            eq0(0(),S(x)) -> 0()
            eq0(S(x),0()) -> 0()
            eq0(S(x'),S(x)) -> eq0(x',x)
        - Signature:
            {eq0/2} / {0/0,S/1}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {eq0} and constructors {0,S}
    + Applied Processor:
        EmptyProcessor
    + Details:
        The problem is already closed. The intended complexity is O(1).

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