YES(O(1),O(n^1))
0.00/0.59	YES(O(1),O(n^1))
0.00/0.59	
0.00/0.59	We are left with following problem, upon which TcT provides the
0.00/0.59	certificate YES(O(1),O(n^1)).
0.00/0.59	
0.00/0.59	Strict Trs:
0.00/0.59	  { f(x, y, s(z)) -> s(f(0(), 1(), z))
0.00/0.59	  , f(0(), 1(), x) -> f(s(x), x, x) }
0.00/0.59	Obligation:
0.00/0.59	  runtime complexity
0.00/0.59	Answer:
0.00/0.59	  YES(O(1),O(n^1))
0.00/0.59	
0.00/0.59	We use the processor 'matrix interpretation of dimension 1' to
0.00/0.59	orient following rules strictly.
0.00/0.59	
0.00/0.59	Trs: { f(x, y, s(z)) -> s(f(0(), 1(), z)) }
0.00/0.59	
0.00/0.59	The induced complexity on above rules (modulo remaining rules) is
0.00/0.59	YES(?,O(n^1)) . These rules are moved into the corresponding weak
0.00/0.59	component(s).
0.00/0.59	
0.00/0.59	Sub-proof:
0.00/0.59	----------
0.00/0.59	  The following argument positions are usable:
0.00/0.59	    Uargs(s) = {1}
0.00/0.59	  
0.00/0.59	  TcT has computed the following constructor-based matrix
0.00/0.59	  interpretation satisfying not(EDA).
0.00/0.59	  
0.00/0.59	    [f](x1, x2, x3) = [4] x3 + [1]
0.00/0.59	                                  
0.00/0.59	                [0] = [0]         
0.00/0.59	                                  
0.00/0.59	                [1] = [0]         
0.00/0.59	                                  
0.00/0.59	            [s](x1) = [1] x1 + [2]
0.00/0.59	  
0.00/0.59	  The order satisfies the following ordering constraints:
0.00/0.59	  
0.00/0.59	     [f(x, y, s(z))] =  [4] z + [9]        
0.00/0.59	                     >  [4] z + [3]        
0.00/0.59	                     =  [s(f(0(), 1(), z))]
0.00/0.59	                                           
0.00/0.59	    [f(0(), 1(), x)] =  [4] x + [1]        
0.00/0.59	                     >= [4] x + [1]        
0.00/0.59	                     =  [f(s(x), x, x)]    
0.00/0.59	                                           
0.00/0.59	
0.00/0.59	We return to the main proof.
0.00/0.59	
0.00/0.59	We are left with following problem, upon which TcT provides the
0.00/0.59	certificate YES(O(1),O(n^1)).
0.00/0.59	
0.00/0.59	Strict Trs: { f(0(), 1(), x) -> f(s(x), x, x) }
0.00/0.59	Weak Trs: { f(x, y, s(z)) -> s(f(0(), 1(), z)) }
0.00/0.59	Obligation:
0.00/0.59	  runtime complexity
0.00/0.59	Answer:
0.00/0.59	  YES(O(1),O(n^1))
0.00/0.59	
0.00/0.59	We use the processor 'matrix interpretation of dimension 2' to
0.00/0.59	orient following rules strictly.
0.00/0.59	
0.00/0.59	Trs: { f(0(), 1(), x) -> f(s(x), x, x) }
0.00/0.59	
0.00/0.59	The induced complexity on above rules (modulo remaining rules) is
0.00/0.59	YES(?,O(n^1)) . These rules are moved into the corresponding weak
0.00/0.59	component(s).
0.00/0.59	
0.00/0.59	Sub-proof:
0.00/0.59	----------
0.00/0.59	  The following argument positions are usable:
0.00/0.59	    Uargs(s) = {1}
0.00/0.59	  
0.00/0.59	  TcT has computed the following constructor-based matrix
0.00/0.59	  interpretation satisfying not(EDA) and not(IDA(1)).
0.00/0.59	  
0.00/0.59	    [f](x1, x2, x3) = [0 2] x1 + [2 0] x3 + [6]
0.00/0.59	                      [0 4]      [0 0]      [4]
0.00/0.59	                                               
0.00/0.59	                [0] = [0]                      
0.00/0.59	                      [1]                      
0.00/0.59	                                               
0.00/0.59	                [1] = [0]                      
0.00/0.59	                      [0]                      
0.00/0.59	                                               
0.00/0.59	            [s](x1) = [1 0] x1 + [2]           
0.00/0.59	                      [0 0]      [0]           
0.00/0.59	  
0.00/0.59	  The order satisfies the following ordering constraints:
0.00/0.59	  
0.00/0.59	     [f(x, y, s(z))] =  [0 2] x + [2 0] z + [10]
0.00/0.59	                        [0 4]     [0 0]     [4] 
0.00/0.59	                     >= [2 0] z + [10]          
0.00/0.59	                        [0 0]     [0]           
0.00/0.59	                     =  [s(f(0(), 1(), z))]     
0.00/0.59	                                                
0.00/0.59	    [f(0(), 1(), x)] =  [2 0] x + [8]           
0.00/0.59	                        [0 0]     [8]           
0.00/0.59	                     >  [2 0] x + [6]           
0.00/0.59	                        [0 0]     [4]           
0.00/0.59	                     =  [f(s(x), x, x)]         
0.00/0.59	                                                
0.00/0.59	
0.00/0.59	We return to the main proof.
0.00/0.59	
0.00/0.59	We are left with following problem, upon which TcT provides the
0.00/0.59	certificate YES(O(1),O(1)).
0.00/0.59	
0.00/0.59	Weak Trs:
0.00/0.59	  { f(x, y, s(z)) -> s(f(0(), 1(), z))
0.00/0.59	  , f(0(), 1(), x) -> f(s(x), x, x) }
0.00/0.59	Obligation:
0.00/0.59	  runtime complexity
0.00/0.59	Answer:
0.00/0.59	  YES(O(1),O(1))
0.00/0.59	
0.00/0.59	Empty rules are trivially bounded
0.00/0.59	
0.00/0.59	Hurray, we answered YES(O(1),O(n^1))
0.00/0.59	EOF