WORST_CASE(?,O(n^3))
* Step 1: NaturalMI WORST_CASE(?,O(n^3))
    + Considered Problem:
        - Strict TRS:
            ifMinus(false(),s(X),Y) -> s(minus(X,Y))
            ifMinus(true(),s(X),Y) -> 0()
            le(0(),Y) -> true()
            le(s(X),0()) -> false()
            le(s(X),s(Y)) -> le(X,Y)
            minus(0(),Y) -> 0()
            minus(s(X),Y) -> ifMinus(le(s(X),Y),s(X),Y)
            quot(0(),s(Y)) -> 0()
            quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
        - Signature:
            {ifMinus/3,le/2,minus/2,quot/2} / {0/0,false/0,s/1,true/0}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {ifMinus,le,minus,quot} and constructors {0,false,s,true}
    + Applied Processor:
        NaturalMI {miDimension = 3, miDegree = 3, miKind = Algebraic, uargs = UArgs, urules = URules, selector = Nothing}
    + Details:
        We apply a matrix interpretation of kind constructor based matrix interpretation:
        The following argument positions are considered usable:
          uargs(ifMinus) = {1},
          uargs(quot) = {1},
          uargs(s) = {1}
        
        Following symbols are considered usable:
          {ifMinus,le,minus,quot}
        TcT has computed the following interpretation:
                p(0) = [0]                          
                       [0]                          
                       [1]                          
            p(false) = [0]                          
                       [0]                          
                       [0]                          
          p(ifMinus) = [1 0 0]      [1 1 1]      [0]
                       [0 0 0] x1 + [0 1 0] x2 + [0]
                       [0 0 0]      [0 0 1]      [0]
               p(le) = [0 0 1]      [0]             
                       [0 0 0] x1 + [0]             
                       [0 0 0]      [0]             
            p(minus) = [1 1 2]      [1]             
                       [0 1 0] x1 + [0]             
                       [0 0 1]      [0]             
             p(quot) = [2 3 2]      [2]             
                       [0 1 0] x1 + [0]             
                       [0 0 1]      [0]             
                p(s) = [1 2 1]      [0]             
                       [0 1 1] x1 + [0]             
                       [0 0 1]      [2]             
             p(true) = [0]                          
                       [0]                          
                       [0]                          
        
        Following rules are strictly oriented:
        ifMinus(false(),s(X),Y) = [1 3 3]     [2]           
                                  [0 1 1] X + [0]           
                                  [0 0 1]     [2]           
                                > [1 3 3]     [1]           
                                  [0 1 1] X + [0]           
                                  [0 0 1]     [2]           
                                = s(minus(X,Y))             
        
         ifMinus(true(),s(X),Y) = [1 3 3]     [2]           
                                  [0 1 1] X + [0]           
                                  [0 0 1]     [2]           
                                > [0]                       
                                  [0]                       
                                  [1]                       
                                = 0()                       
        
                      le(0(),Y) = [1]                       
                                  [0]                       
                                  [0]                       
                                > [0]                       
                                  [0]                       
                                  [0]                       
                                = true()                    
        
                   le(s(X),0()) = [0 0 1]     [2]           
                                  [0 0 0] X + [0]           
                                  [0 0 0]     [0]           
                                > [0]                       
                                  [0]                       
                                  [0]                       
                                = false()                   
        
                  le(s(X),s(Y)) = [0 0 1]     [2]           
                                  [0 0 0] X + [0]           
                                  [0 0 0]     [0]           
                                > [0 0 1]     [0]           
                                  [0 0 0] X + [0]           
                                  [0 0 0]     [0]           
                                = le(X,Y)                   
        
                   minus(0(),Y) = [3]                       
                                  [0]                       
                                  [1]                       
                                > [0]                       
                                  [0]                       
                                  [1]                       
                                = 0()                       
        
                  minus(s(X),Y) = [1 3 4]     [5]           
                                  [0 1 1] X + [0]           
                                  [0 0 1]     [2]           
                                > [1 3 4]     [4]           
                                  [0 1 1] X + [0]           
                                  [0 0 1]     [2]           
                                = ifMinus(le(s(X),Y),s(X),Y)
        
                 quot(0(),s(Y)) = [4]                       
                                  [0]                       
                                  [1]                       
                                > [0]                       
                                  [0]                       
                                  [1]                       
                                = 0()                       
        
                quot(s(X),s(Y)) = [2 7 7]     [6]           
                                  [0 1 1] X + [0]           
                                  [0 0 1]     [2]           
                                > [2 7 7]     [4]           
                                  [0 1 1] X + [0]           
                                  [0 0 1]     [2]           
                                = s(quot(minus(X,Y),s(Y)))  
        
        
        Following rules are (at-least) weakly oriented:
        

WORST_CASE(?,O(n^3))