YES(O(1),O(n^1))
0.00/0.25	YES(O(1),O(n^1))
0.00/0.25	
0.00/0.25	We are left with following problem, upon which TcT provides the
0.00/0.25	certificate YES(O(1),O(n^1)).
0.00/0.25	
0.00/0.25	Strict Trs:
0.00/0.25	  { f(0(), y) -> 0()
0.00/0.25	  , f(s(x), y) -> f(f(x, y), y) }
0.00/0.25	Obligation:
0.00/0.25	  runtime complexity
0.00/0.25	Answer:
0.00/0.25	  YES(O(1),O(n^1))
0.00/0.25	
0.00/0.25	We add the following weak dependency pairs:
0.00/0.25	
0.00/0.25	Strict DPs:
0.00/0.25	  { f^#(0(), y) -> c_1()
0.00/0.25	  , f^#(s(x), y) -> c_2(f^#(f(x, y), y)) }
0.00/0.25	
0.00/0.25	and mark the set of starting terms.
0.00/0.25	
0.00/0.25	We are left with following problem, upon which TcT provides the
0.00/0.25	certificate YES(O(1),O(n^1)).
0.00/0.25	
0.00/0.25	Strict DPs:
0.00/0.25	  { f^#(0(), y) -> c_1()
0.00/0.25	  , f^#(s(x), y) -> c_2(f^#(f(x, y), y)) }
0.00/0.25	Strict Trs:
0.00/0.25	  { f(0(), y) -> 0()
0.00/0.25	  , f(s(x), y) -> f(f(x, y), y) }
0.00/0.25	Obligation:
0.00/0.25	  runtime complexity
0.00/0.25	Answer:
0.00/0.25	  YES(O(1),O(n^1))
0.00/0.25	
0.00/0.25	The weightgap principle applies (using the following constant
0.00/0.25	growth matrix-interpretation)
0.00/0.25	
0.00/0.25	The following argument positions are usable:
0.00/0.25	  Uargs(f) = {1}, Uargs(f^#) = {1}, Uargs(c_2) = {1}
0.00/0.25	
0.00/0.25	TcT has computed the following constructor-restricted matrix
0.00/0.25	interpretation.
0.00/0.25	
0.00/0.25	    [f](x1, x2) = [1 1] x1 + [0]           
0.00/0.25	                  [0 0]      [1]           
0.00/0.25	                                           
0.00/0.25	            [0] = [0]                      
0.00/0.25	                  [1]                      
0.00/0.25	                                           
0.00/0.25	        [s](x1) = [1 0] x1 + [2]           
0.00/0.25	                  [0 1]      [2]           
0.00/0.25	                                           
0.00/0.25	  [f^#](x1, x2) = [1 1] x1 + [2 2] x2 + [0]
0.00/0.25	                  [0 0]      [2 2]      [0]
0.00/0.25	                                           
0.00/0.25	          [c_1] = [1]                      
0.00/0.25	                  [2]                      
0.00/0.25	                                           
0.00/0.25	      [c_2](x1) = [1 0] x1 + [1]           
0.00/0.25	                  [0 1]      [2]           
0.00/0.25	
0.00/0.25	The order satisfies the following ordering constraints:
0.00/0.25	
0.00/0.25	     [f(0(), y)] = [1]                    
0.00/0.25	                   [1]                    
0.00/0.25	                 > [0]                    
0.00/0.25	                   [1]                    
0.00/0.25	                 = [0()]                  
0.00/0.25	                                          
0.00/0.25	    [f(s(x), y)] = [1 1] x + [4]          
0.00/0.25	                   [0 0]     [1]          
0.00/0.25	                 > [1 1] x + [1]          
0.00/0.25	                   [0 0]     [1]          
0.00/0.25	                 = [f(f(x, y), y)]        
0.00/0.25	                                          
0.00/0.25	   [f^#(0(), y)] = [2 2] y + [1]          
0.00/0.25	                   [2 2]     [0]          
0.00/0.25	                 ? [1]                    
0.00/0.25	                   [2]                    
0.00/0.25	                 = [c_1()]                
0.00/0.25	                                          
0.00/0.25	  [f^#(s(x), y)] = [2 2] y + [1 1] x + [4]
0.00/0.25	                   [2 2]     [0 0]     [0]
0.00/0.25	                 ? [2 2] y + [1 1] x + [2]
0.00/0.25	                   [2 2]     [0 0]     [2]
0.00/0.25	                 = [c_2(f^#(f(x, y), y))] 
0.00/0.25	                                          
0.00/0.25	
0.00/0.25	Further, it can be verified that all rules not oriented are covered by the weightgap condition.
0.00/0.25	
0.00/0.25	We are left with following problem, upon which TcT provides the
0.00/0.25	certificate YES(?,O(n^1)).
0.00/0.25	
0.00/0.25	Strict DPs:
0.00/0.25	  { f^#(0(), y) -> c_1()
0.00/0.25	  , f^#(s(x), y) -> c_2(f^#(f(x, y), y)) }
0.00/0.25	Weak Trs:
0.00/0.25	  { f(0(), y) -> 0()
0.00/0.25	  , f(s(x), y) -> f(f(x, y), y) }
0.00/0.25	Obligation:
0.00/0.25	  runtime complexity
0.00/0.25	Answer:
0.00/0.25	  YES(?,O(n^1))
0.00/0.25	
0.00/0.25	We estimate the number of application of {1} by applications of
0.00/0.25	Pre({1}) = {2}. Here rules are labeled as follows:
0.00/0.25	
0.00/0.25	  DPs:
0.00/0.25	    { 1: f^#(0(), y) -> c_1()
0.00/0.25	    , 2: f^#(s(x), y) -> c_2(f^#(f(x, y), y)) }
0.00/0.25	
0.00/0.25	We are left with following problem, upon which TcT provides the
0.00/0.25	certificate YES(?,O(n^1)).
0.00/0.25	
0.00/0.25	Strict DPs: { f^#(s(x), y) -> c_2(f^#(f(x, y), y)) }
0.00/0.25	Weak DPs: { f^#(0(), y) -> c_1() }
0.00/0.25	Weak Trs:
0.00/0.25	  { f(0(), y) -> 0()
0.00/0.25	  , f(s(x), y) -> f(f(x, y), y) }
0.00/0.25	Obligation:
0.00/0.25	  runtime complexity
0.00/0.25	Answer:
0.00/0.25	  YES(?,O(n^1))
0.00/0.25	
0.00/0.25	We estimate the number of application of {1} by applications of
0.00/0.25	Pre({1}) = {}. Here rules are labeled as follows:
0.00/0.25	
0.00/0.25	  DPs:
0.00/0.25	    { 1: f^#(s(x), y) -> c_2(f^#(f(x, y), y))
0.00/0.25	    , 2: f^#(0(), y) -> c_1() }
0.00/0.25	
0.00/0.25	We are left with following problem, upon which TcT provides the
0.00/0.25	certificate YES(?,O(n^1)).
0.00/0.25	
0.00/0.25	Weak DPs:
0.00/0.25	  { f^#(0(), y) -> c_1()
0.00/0.25	  , f^#(s(x), y) -> c_2(f^#(f(x, y), y)) }
0.00/0.25	Weak Trs:
0.00/0.25	  { f(0(), y) -> 0()
0.00/0.25	  , f(s(x), y) -> f(f(x, y), y) }
0.00/0.25	Obligation:
0.00/0.25	  runtime complexity
0.00/0.25	Answer:
0.00/0.25	  YES(?,O(n^1))
0.00/0.25	
0.00/0.25	The following weak DPs constitute a sub-graph of the DG that is
0.00/0.25	closed under successors. The DPs are removed.
0.00/0.25	
0.00/0.25	{ f^#(0(), y) -> c_1()
0.00/0.25	, f^#(s(x), y) -> c_2(f^#(f(x, y), y)) }
0.00/0.25	
0.00/0.25	We are left with following problem, upon which TcT provides the
0.00/0.25	certificate YES(?,O(n^1)).
0.00/0.25	
0.00/0.25	Weak Trs:
0.00/0.25	  { f(0(), y) -> 0()
0.00/0.25	  , f(s(x), y) -> f(f(x, y), y) }
0.00/0.25	Obligation:
0.00/0.25	  runtime complexity
0.00/0.25	Answer:
0.00/0.25	  YES(?,O(n^1))
0.00/0.25	
0.00/0.25	We employ 'linear path analysis' using the following approximated
0.00/0.25	dependency graph:
0.00/0.25	empty
0.00/0.25	
0.00/0.25	
0.00/0.25	Hurray, we answered YES(O(1),O(n^1))
0.00/0.25	EOF