YES(O(1),O(n^2))
162.50/60.05	YES(O(1),O(n^2))
162.50/60.05	
162.50/60.05	We are left with following problem, upon which TcT provides the
162.50/60.05	certificate YES(O(1),O(n^2)).
162.50/60.05	
162.50/60.05	Strict Trs:
162.50/60.05	  { f(a, empty()) -> g(a, empty())
162.50/60.05	  , f(a, cons(x, k)) -> f(cons(x, a), k)
162.50/60.05	  , g(empty(), d) -> d
162.50/60.05	  , g(cons(x, k), d) -> g(k, cons(x, d)) }
162.50/60.05	Obligation:
162.50/60.05	  derivational complexity
162.50/60.05	Answer:
162.50/60.05	  YES(O(1),O(n^2))
162.50/60.05	
162.50/60.05	The weightgap principle applies (using the following nonconstant
162.50/60.05	growth matrix-interpretation)
162.50/60.05	
162.50/60.05	TcT has computed the following triangular matrix interpretation.
162.50/60.05	Note that the diagonal of the component-wise maxima of
162.50/60.05	interpretation-entries contains no more than 1 non-zero entries.
162.50/60.05	
162.50/60.05	     [f](x1, x2) = [1] x1 + [1] x2 + [0]
162.50/60.05	                                        
162.50/60.05	         [empty] = [0]                  
162.50/60.05	                                        
162.50/60.05	     [g](x1, x2) = [1] x1 + [1] x2 + [1]
162.50/60.05	                                        
162.50/60.05	  [cons](x1, x2) = [1] x1 + [1] x2 + [0]
162.50/60.05	
162.50/60.05	The order satisfies the following ordering constraints:
162.50/60.05	
162.50/60.05	     [f(a, empty())] =  [1] a + [0]                
162.50/60.05	                     ?  [1] a + [1]                
162.50/60.05	                     =  [g(a, empty())]            
162.50/60.05	                                                   
162.50/60.05	  [f(a, cons(x, k))] =  [1] a + [1] x + [1] k + [0]
162.50/60.05	                     >= [1] a + [1] x + [1] k + [0]
162.50/60.05	                     =  [f(cons(x, a), k)]         
162.50/60.05	                                                   
162.50/60.05	     [g(empty(), d)] =  [1] d + [1]                
162.50/60.05	                     >  [1] d + [0]                
162.50/60.05	                     =  [d]                        
162.50/60.05	                                                   
162.50/60.05	  [g(cons(x, k), d)] =  [1] x + [1] k + [1] d + [1]
162.50/60.05	                     >= [1] x + [1] k + [1] d + [1]
162.50/60.05	                     =  [g(k, cons(x, d))]         
162.50/60.05	                                                   
162.50/60.05	
162.50/60.05	Further, it can be verified that all rules not oriented are covered by the weightgap condition.
162.50/60.05	
162.50/60.05	We are left with following problem, upon which TcT provides the
162.50/60.05	certificate YES(O(1),O(n^2)).
162.50/60.05	
162.50/60.05	Strict Trs:
162.50/60.05	  { f(a, empty()) -> g(a, empty())
162.50/60.05	  , f(a, cons(x, k)) -> f(cons(x, a), k)
162.50/60.05	  , g(cons(x, k), d) -> g(k, cons(x, d)) }
162.50/60.05	Weak Trs: { g(empty(), d) -> d }
162.50/60.05	Obligation:
162.50/60.05	  derivational complexity
162.50/60.05	Answer:
162.50/60.05	  YES(O(1),O(n^2))
162.50/60.05	
162.50/60.05	We use the processor 'matrix interpretation of dimension 1' to
162.50/60.05	orient following rules strictly.
162.50/60.05	
162.50/60.05	Trs: { f(a, empty()) -> g(a, empty()) }
162.50/60.05	
162.50/60.05	The induced complexity on above rules (modulo remaining rules) is
162.50/60.05	YES(?,O(n^1)) . These rules are moved into the corresponding weak
162.50/60.05	component(s).
162.50/60.05	
162.50/60.05	Sub-proof:
162.50/60.05	----------
162.50/60.05	  TcT has computed the following triangular matrix interpretation.
162.50/60.05	  
162.50/60.05	       [f](x1, x2) = [1] x1 + [1] x2 + [2]
162.50/60.05	                                          
162.50/60.05	           [empty] = [2]                  
162.50/60.05	                                          
162.50/60.05	       [g](x1, x2) = [1] x1 + [1] x2 + [1]
162.50/60.05	                                          
162.50/60.05	    [cons](x1, x2) = [1] x1 + [1] x2 + [2]
162.50/60.05	  
162.50/60.05	  The order satisfies the following ordering constraints:
162.50/60.05	  
162.50/60.05	       [f(a, empty())] =  [1] a + [4]                
162.50/60.05	                       >  [1] a + [3]                
162.50/60.05	                       =  [g(a, empty())]            
162.50/60.05	                                                     
162.50/60.05	    [f(a, cons(x, k))] =  [1] a + [1] x + [1] k + [4]
162.50/60.05	                       >= [1] a + [1] x + [1] k + [4]
162.50/60.05	                       =  [f(cons(x, a), k)]         
162.50/60.05	                                                     
162.50/60.05	       [g(empty(), d)] =  [1] d + [3]                
162.50/60.05	                       >  [1] d + [0]                
162.50/60.05	                       =  [d]                        
162.50/60.05	                                                     
162.50/60.05	    [g(cons(x, k), d)] =  [1] x + [1] k + [1] d + [3]
162.50/60.05	                       >= [1] x + [1] k + [1] d + [3]
162.50/60.05	                       =  [g(k, cons(x, d))]         
162.50/60.05	                                                     
162.50/60.05	
162.50/60.05	We return to the main proof.
162.50/60.05	
162.50/60.05	We are left with following problem, upon which TcT provides the
162.50/60.05	certificate YES(O(1),O(n^2)).
162.50/60.05	
162.50/60.05	Strict Trs:
162.50/60.05	  { f(a, cons(x, k)) -> f(cons(x, a), k)
162.50/60.05	  , g(cons(x, k), d) -> g(k, cons(x, d)) }
162.50/60.05	Weak Trs:
162.50/60.05	  { f(a, empty()) -> g(a, empty())
162.50/60.05	  , g(empty(), d) -> d }
162.50/60.05	Obligation:
162.50/60.05	  derivational complexity
162.50/60.05	Answer:
162.50/60.05	  YES(O(1),O(n^2))
162.50/60.05	
162.50/60.05	We use the processor 'matrix interpretation of dimension 2' to
162.50/60.05	orient following rules strictly.
162.50/60.05	
162.50/60.05	Trs: { f(a, cons(x, k)) -> f(cons(x, a), k) }
162.50/60.05	
162.50/60.05	The induced complexity on above rules (modulo remaining rules) is
162.50/60.05	YES(?,O(n^2)) . These rules are moved into the corresponding weak
162.50/60.05	component(s).
162.50/60.05	
162.50/60.05	Sub-proof:
162.50/60.05	----------
162.50/60.05	  TcT has computed the following triangular matrix interpretation.
162.50/60.05	  
162.50/60.05	       [f](x1, x2) = [1 0] x1 + [1 1] x2 + [0]
162.50/60.05	                     [0 1]      [0 1]      [0]
162.50/60.05	                                              
162.50/60.05	           [empty] = [0]                      
162.50/60.05	                     [0]                      
162.50/60.05	                                              
162.50/60.05	       [g](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
162.50/60.05	                     [0 1]      [0 1]      [0]
162.50/60.05	                                              
162.50/60.05	    [cons](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
162.50/60.05	                     [0 0]      [0 1]      [1]
162.50/60.05	  
162.50/60.05	  The order satisfies the following ordering constraints:
162.50/60.05	  
162.50/60.05	       [f(a, empty())] =  [1 0] a + [0]                    
162.50/60.05	                          [0 1]     [0]                    
162.50/60.05	                       >= [1 0] a + [0]                    
162.50/60.05	                          [0 1]     [0]                    
162.50/60.05	                       =  [g(a, empty())]                  
162.50/60.05	                                                           
162.50/60.05	    [f(a, cons(x, k))] =  [1 0] a + [1 0] x + [1 1] k + [1]
162.50/60.05	                          [0 1]     [0 0]     [0 1]     [1]
162.50/60.05	                       >  [1 0] a + [1 0] x + [1 1] k + [0]
162.50/60.05	                          [0 1]     [0 0]     [0 1]     [1]
162.50/60.05	                       =  [f(cons(x, a), k)]               
162.50/60.05	                                                           
162.50/60.05	       [g(empty(), d)] =  [1 0] d + [0]                    
162.50/60.05	                          [0 1]     [0]                    
162.50/60.05	                       >= [1 0] d + [0]                    
162.50/60.05	                          [0 1]     [0]                    
162.50/60.05	                       =  [d]                              
162.50/60.05	                                                           
162.50/60.05	    [g(cons(x, k), d)] =  [1 0] x + [1 0] k + [1 0] d + [0]
162.50/60.05	                          [0 0]     [0 1]     [0 1]     [1]
162.50/60.05	                       >= [1 0] x + [1 0] k + [1 0] d + [0]
162.50/60.05	                          [0 0]     [0 1]     [0 1]     [1]
162.50/60.05	                       =  [g(k, cons(x, d))]               
162.50/60.05	                                                           
162.50/60.05	
162.50/60.05	We return to the main proof.
162.50/60.05	
162.50/60.05	We are left with following problem, upon which TcT provides the
162.50/60.05	certificate YES(O(1),O(n^2)).
162.50/60.05	
162.50/60.05	Strict Trs: { g(cons(x, k), d) -> g(k, cons(x, d)) }
162.50/60.05	Weak Trs:
162.50/60.05	  { f(a, empty()) -> g(a, empty())
162.50/60.05	  , f(a, cons(x, k)) -> f(cons(x, a), k)
162.50/60.05	  , g(empty(), d) -> d }
162.50/60.05	Obligation:
162.50/60.05	  derivational complexity
162.50/60.05	Answer:
162.50/60.05	  YES(O(1),O(n^2))
162.50/60.05	
162.50/60.05	We use the processor 'matrix interpretation of dimension 2' to
162.50/60.05	orient following rules strictly.
162.50/60.05	
162.50/60.05	Trs: { g(cons(x, k), d) -> g(k, cons(x, d)) }
162.50/60.05	
162.50/60.05	The induced complexity on above rules (modulo remaining rules) is
162.50/60.05	YES(?,O(n^2)) . These rules are moved into the corresponding weak
162.50/60.05	component(s).
162.50/60.05	
162.50/60.05	Sub-proof:
162.50/60.05	----------
162.50/60.05	  TcT has computed the following triangular matrix interpretation.
162.50/60.05	  
162.50/60.05	       [f](x1, x2) = [1 1] x1 + [1 1] x2 + [0]
162.50/60.05	                     [0 1]      [0 1]      [0]
162.50/60.05	                                              
162.50/60.05	           [empty] = [0]                      
162.50/60.05	                     [0]                      
162.50/60.05	                                              
162.50/60.05	       [g](x1, x2) = [1 1] x1 + [1 0] x2 + [0]
162.50/60.05	                     [0 1]      [0 1]      [0]
162.50/60.05	                                              
162.50/60.05	    [cons](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
162.50/60.05	                     [0 0]      [0 1]      [1]
162.50/60.05	  
162.50/60.05	  The order satisfies the following ordering constraints:
162.50/60.05	  
162.50/60.05	       [f(a, empty())] =  [1 1] a + [0]                    
162.50/60.05	                          [0 1]     [0]                    
162.50/60.05	                       >= [1 1] a + [0]                    
162.50/60.05	                          [0 1]     [0]                    
162.50/60.05	                       =  [g(a, empty())]                  
162.50/60.05	                                                           
162.50/60.05	    [f(a, cons(x, k))] =  [1 1] a + [1 0] x + [1 1] k + [1]
162.50/60.05	                          [0 1]     [0 0]     [0 1]     [1]
162.50/60.05	                       >= [1 1] a + [1 0] x + [1 1] k + [1]
162.50/60.05	                          [0 1]     [0 0]     [0 1]     [1]
162.50/60.05	                       =  [f(cons(x, a), k)]               
162.50/60.05	                                                           
162.50/60.05	       [g(empty(), d)] =  [1 0] d + [0]                    
162.50/60.05	                          [0 1]     [0]                    
162.50/60.05	                       >= [1 0] d + [0]                    
162.50/60.05	                          [0 1]     [0]                    
162.50/60.05	                       =  [d]                              
162.50/60.05	                                                           
162.50/60.05	    [g(cons(x, k), d)] =  [1 0] x + [1 1] k + [1 0] d + [1]
162.50/60.05	                          [0 0]     [0 1]     [0 1]     [1]
162.50/60.05	                       >  [1 0] x + [1 1] k + [1 0] d + [0]
162.50/60.05	                          [0 0]     [0 1]     [0 1]     [1]
162.50/60.05	                       =  [g(k, cons(x, d))]               
162.50/60.05	                                                           
162.50/60.05	
162.50/60.05	We return to the main proof.
162.50/60.05	
162.50/60.05	We are left with following problem, upon which TcT provides the
162.50/60.05	certificate YES(O(1),O(1)).
162.50/60.05	
162.50/60.05	Weak Trs:
162.50/60.05	  { f(a, empty()) -> g(a, empty())
162.50/60.05	  , f(a, cons(x, k)) -> f(cons(x, a), k)
162.50/60.05	  , g(empty(), d) -> d
162.50/60.05	  , g(cons(x, k), d) -> g(k, cons(x, d)) }
162.50/60.05	Obligation:
162.50/60.05	  derivational complexity
162.50/60.05	Answer:
162.50/60.05	  YES(O(1),O(1))
162.50/60.05	
162.50/60.05	Empty rules are trivially bounded
162.50/60.05	
162.50/60.05	Hurray, we answered YES(O(1),O(n^2))
162.50/60.05	EOF