YES(O(1),O(n^1))
3.12/1.25	YES(O(1),O(n^1))
3.12/1.25	
3.12/1.25	We are left with following problem, upon which TcT provides the
3.12/1.25	certificate YES(O(1),O(n^1)).
3.12/1.25	
3.12/1.25	Strict Trs:
3.12/1.25	  { perfectp(0()) -> false()
3.12/1.25	  , perfectp(s(x)) -> f(x, s(0()), s(x), s(x))
3.12/1.25	  , f(0(), y, 0(), u) -> true()
3.12/1.25	  , f(0(), y, s(z), u) -> false()
3.12/1.25	  , f(s(x), 0(), z, u) -> f(x, u, minus(z, s(x)), u)
3.12/1.25	  , f(s(x), s(y), z, u) ->
3.12/1.25	    if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) }
3.12/1.25	Obligation:
3.12/1.25	  runtime complexity
3.12/1.25	Answer:
3.12/1.25	  YES(O(1),O(n^1))
3.12/1.25	
3.12/1.25	We add the following weak dependency pairs:
3.12/1.25	
3.12/1.25	Strict DPs:
3.12/1.25	  { perfectp^#(0()) -> c_1()
3.12/1.25	  , perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
3.12/1.25	  , f^#(0(), y, 0(), u) -> c_3()
3.12/1.25	  , f^#(0(), y, s(z), u) -> c_4()
3.12/1.25	  , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
3.12/1.25	  , f^#(s(x), s(y), z, u) ->
3.12/1.25	    c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
3.12/1.25	
3.12/1.25	and mark the set of starting terms.
3.12/1.25	
3.12/1.25	We are left with following problem, upon which TcT provides the
3.12/1.25	certificate YES(O(1),O(n^1)).
3.12/1.25	
3.12/1.25	Strict DPs:
3.12/1.25	  { perfectp^#(0()) -> c_1()
3.12/1.25	  , perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
3.12/1.25	  , f^#(0(), y, 0(), u) -> c_3()
3.12/1.25	  , f^#(0(), y, s(z), u) -> c_4()
3.12/1.25	  , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
3.12/1.25	  , f^#(s(x), s(y), z, u) ->
3.12/1.25	    c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
3.12/1.25	Strict Trs:
3.12/1.25	  { perfectp(0()) -> false()
3.12/1.25	  , perfectp(s(x)) -> f(x, s(0()), s(x), s(x))
3.12/1.25	  , f(0(), y, 0(), u) -> true()
3.12/1.25	  , f(0(), y, s(z), u) -> false()
3.12/1.25	  , f(s(x), 0(), z, u) -> f(x, u, minus(z, s(x)), u)
3.12/1.25	  , f(s(x), s(y), z, u) ->
3.12/1.25	    if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) }
3.12/1.25	Obligation:
3.12/1.25	  runtime complexity
3.12/1.25	Answer:
3.12/1.25	  YES(O(1),O(n^1))
3.12/1.25	
3.12/1.25	No rule is usable, rules are removed from the input problem.
3.12/1.25	
3.12/1.25	We are left with following problem, upon which TcT provides the
3.12/1.25	certificate YES(O(1),O(n^1)).
3.12/1.25	
3.12/1.25	Strict DPs:
3.12/1.25	  { perfectp^#(0()) -> c_1()
3.12/1.25	  , perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
3.12/1.25	  , f^#(0(), y, 0(), u) -> c_3()
3.12/1.25	  , f^#(0(), y, s(z), u) -> c_4()
3.12/1.25	  , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
3.12/1.25	  , f^#(s(x), s(y), z, u) ->
3.12/1.25	    c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
3.12/1.25	Obligation:
3.12/1.25	  runtime complexity
3.12/1.25	Answer:
3.12/1.25	  YES(O(1),O(n^1))
3.12/1.25	
3.12/1.25	The weightgap principle applies (using the following constant
3.12/1.25	growth matrix-interpretation)
3.12/1.25	
3.12/1.25	The following argument positions are usable:
3.12/1.25	  Uargs(c_2) = {1}, Uargs(c_5) = {1}, Uargs(c_6) = {4}
3.12/1.25	
3.12/1.25	TcT has computed the following constructor-restricted matrix
3.12/1.25	interpretation.
3.12/1.25	
3.12/1.25	                    [0] = [0]                                 
3.12/1.25	                          [0]                                 
3.12/1.25	                                                              
3.12/1.25	                [s](x1) = [0]                                 
3.12/1.25	                          [0]                                 
3.12/1.25	                                                              
3.12/1.25	        [minus](x1, x2) = [1 0] x1 + [0]                      
3.12/1.25	                          [0 0]      [0]                      
3.12/1.25	                                                              
3.12/1.25	       [perfectp^#](x1) = [0]                                 
3.12/1.25	                          [0]                                 
3.12/1.25	                                                              
3.12/1.25	                  [c_1] = [1]                                 
3.12/1.25	                          [0]                                 
3.12/1.25	                                                              
3.12/1.25	              [c_2](x1) = [1 0] x1 + [0]                      
3.12/1.25	                          [0 1]      [0]                      
3.12/1.25	                                                              
3.12/1.25	  [f^#](x1, x2, x3, x4) = [1]                                 
3.12/1.25	                          [0]                                 
3.12/1.25	                                                              
3.12/1.25	                  [c_3] = [0]                                 
3.12/1.25	                          [0]                                 
3.12/1.25	                                                              
3.12/1.25	                  [c_4] = [0]                                 
3.12/1.25	                          [0]                                 
3.12/1.25	                                                              
3.12/1.25	              [c_5](x1) = [1 0] x1 + [0]                      
3.12/1.25	                          [0 1]      [0]                      
3.12/1.25	                                                              
3.12/1.25	  [c_6](x1, x2, x3, x4) = [0 0] x1 + [0 0] x2 + [1 0] x4 + [0]
3.12/1.25	                          [1 2]      [2 0]      [0 1]      [0]
3.12/1.25	
3.12/1.25	The order satisfies the following ordering constraints:
3.12/1.25	
3.12/1.25	        [perfectp^#(0())] =  [0]                                                       
3.12/1.25	                             [0]                                                       
3.12/1.25	                          ?  [1]                                                       
3.12/1.25	                             [0]                                                       
3.12/1.25	                          =  [c_1()]                                                   
3.12/1.25	                                                                                       
3.12/1.25	       [perfectp^#(s(x))] =  [0]                                                       
3.12/1.25	                             [0]                                                       
3.12/1.25	                          ?  [1]                                                       
3.12/1.25	                             [0]                                                       
3.12/1.25	                          =  [c_2(f^#(x, s(0()), s(x), s(x)))]                         
3.12/1.25	                                                                                       
3.12/1.25	    [f^#(0(), y, 0(), u)] =  [1]                                                       
3.12/1.25	                             [0]                                                       
3.12/1.25	                          >  [0]                                                       
3.12/1.25	                             [0]                                                       
3.12/1.25	                          =  [c_3()]                                                   
3.12/1.25	                                                                                       
3.12/1.25	   [f^#(0(), y, s(z), u)] =  [1]                                                       
3.12/1.25	                             [0]                                                       
3.12/1.25	                          >  [0]                                                       
3.12/1.25	                             [0]                                                       
3.12/1.25	                          =  [c_4()]                                                   
3.12/1.25	                                                                                       
3.12/1.25	   [f^#(s(x), 0(), z, u)] =  [1]                                                       
3.12/1.25	                             [0]                                                       
3.12/1.25	                          >= [1]                                                       
3.12/1.25	                             [0]                                                       
3.12/1.25	                          =  [c_5(f^#(x, u, minus(z, s(x)), u))]                       
3.12/1.25	                                                                                       
3.12/1.25	  [f^#(s(x), s(y), z, u)] =  [1]                                                       
3.12/1.25	                             [0]                                                       
3.12/1.25	                          ?  [0 0] x + [0 0] y + [1]                                   
3.12/1.25	                             [1 2]     [2 0]     [0]                                   
3.12/1.25	                          =  [c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u))]
3.12/1.25	                                                                                       
3.12/1.25	
3.12/1.25	Further, it can be verified that all rules not oriented are covered by the weightgap condition.
3.12/1.25	
3.12/1.25	We are left with following problem, upon which TcT provides the
3.12/1.25	certificate YES(O(1),O(n^1)).
3.12/1.25	
3.12/1.25	Strict DPs:
3.12/1.25	  { perfectp^#(0()) -> c_1()
3.12/1.25	  , perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
3.12/1.25	  , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
3.12/1.25	  , f^#(s(x), s(y), z, u) ->
3.12/1.25	    c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
3.12/1.25	Weak DPs:
3.12/1.25	  { f^#(0(), y, 0(), u) -> c_3()
3.12/1.25	  , f^#(0(), y, s(z), u) -> c_4() }
3.12/1.25	Obligation:
3.12/1.25	  runtime complexity
3.12/1.25	Answer:
3.12/1.25	  YES(O(1),O(n^1))
3.12/1.25	
3.12/1.25	We estimate the number of application of {1} by applications of
3.12/1.25	Pre({1}) = {4}. Here rules are labeled as follows:
3.12/1.25	
3.12/1.25	  DPs:
3.12/1.25	    { 1: perfectp^#(0()) -> c_1()
3.12/1.25	    , 2: perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
3.12/1.25	    , 3: f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
3.12/1.25	    , 4: f^#(s(x), s(y), z, u) ->
3.12/1.25	         c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u))
3.12/1.25	    , 5: f^#(0(), y, 0(), u) -> c_3()
3.12/1.25	    , 6: f^#(0(), y, s(z), u) -> c_4() }
3.12/1.25	
3.12/1.25	We are left with following problem, upon which TcT provides the
3.12/1.25	certificate YES(O(1),O(n^1)).
3.12/1.25	
3.12/1.25	Strict DPs:
3.12/1.25	  { perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
3.12/1.25	  , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
3.12/1.25	  , f^#(s(x), s(y), z, u) ->
3.12/1.25	    c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
3.12/1.25	Weak DPs:
3.12/1.25	  { perfectp^#(0()) -> c_1()
3.12/1.25	  , f^#(0(), y, 0(), u) -> c_3()
3.12/1.29	  , f^#(0(), y, s(z), u) -> c_4() }
3.12/1.29	Obligation:
3.12/1.29	  runtime complexity
3.12/1.29	Answer:
3.12/1.29	  YES(O(1),O(n^1))
3.12/1.29	
3.12/1.29	The following weak DPs constitute a sub-graph of the DG that is
3.12/1.29	closed under successors. The DPs are removed.
3.12/1.29	
3.12/1.29	{ perfectp^#(0()) -> c_1()
3.12/1.29	, f^#(0(), y, 0(), u) -> c_3()
3.12/1.29	, f^#(0(), y, s(z), u) -> c_4() }
3.12/1.29	
3.12/1.29	We are left with following problem, upon which TcT provides the
3.12/1.29	certificate YES(O(1),O(n^1)).
3.12/1.29	
3.12/1.29	Strict DPs:
3.12/1.29	  { perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x)))
3.12/1.29	  , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u))
3.12/1.29	  , f^#(s(x), s(y), z, u) ->
3.12/1.29	    c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
3.12/1.29	Obligation:
3.12/1.29	  runtime complexity
3.12/1.29	Answer:
3.12/1.29	  YES(O(1),O(n^1))
3.12/1.29	
3.12/1.29	Due to missing edges in the dependency-graph, the right-hand sides
3.12/1.29	of following rules could be simplified:
3.12/1.29	
3.12/1.29	  { f^#(s(x), s(y), z, u) ->
3.12/1.29	    c_6(x, y, f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) }
3.12/1.29	
3.12/1.29	We are left with following problem, upon which TcT provides the
3.12/1.29	certificate YES(O(1),O(n^1)).
3.12/1.29	
3.12/1.29	Strict DPs:
3.12/1.29	  { perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x)))
3.12/1.29	  , f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u))
3.12/1.29	  , f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u)) }
3.12/1.29	Obligation:
3.12/1.29	  runtime complexity
3.12/1.29	Answer:
3.12/1.29	  YES(O(1),O(n^1))
3.12/1.29	
3.12/1.29	We use the processor 'matrix interpretation of dimension 1' to
3.12/1.29	orient following rules strictly.
3.12/1.29	
3.12/1.29	DPs:
3.12/1.29	  { 1: perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) }
3.12/1.29	
3.12/1.29	Sub-proof:
3.12/1.29	----------
3.12/1.29	  The following argument positions are usable:
3.12/1.29	    Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {3}
3.12/1.29	  
3.12/1.29	  TcT has computed the following constructor-based matrix
3.12/1.29	  interpretation satisfying not(EDA).
3.12/1.29	  
3.12/1.29	                      [0] = [0]         
3.12/1.29	                                        
3.12/1.29	                  [s](x1) = [1] x1 + [0]
3.12/1.29	                                        
3.12/1.29	          [minus](x1, x2) = [1] x1 + [4]
3.12/1.29	                                        
3.12/1.29	         [perfectp^#](x1) = [4] x1 + [7]
3.12/1.29	                                        
3.12/1.29	    [f^#](x1, x2, x3, x4) = [0]         
3.12/1.29	                                        
3.12/1.29	                [c_1](x1) = [4] x1 + [1]
3.12/1.29	                                        
3.12/1.29	                [c_2](x1) = [4] x1 + [0]
3.12/1.29	                                        
3.12/1.29	        [c_3](x1, x2, x3) = [2] x3 + [0]
3.12/1.29	  
3.12/1.29	  The order satisfies the following ordering constraints:
3.12/1.29	  
3.12/1.29	         [perfectp^#(s(x))] =  [4] x + [7]                        
3.12/1.29	                            >  [1]                                
3.12/1.29	                            =  [c_1(f^#(x, s(0()), s(x), s(x)))]  
3.12/1.29	                                                                  
3.12/1.29	     [f^#(s(x), 0(), z, u)] =  [0]                                
3.12/1.29	                            >= [0]                                
3.12/1.29	                            =  [c_2(f^#(x, u, minus(z, s(x)), u))]
3.12/1.29	                                                                  
3.12/1.29	    [f^#(s(x), s(y), z, u)] =  [0]                                
3.12/1.29	                            >= [0]                                
3.12/1.29	                            =  [c_3(x, y, f^#(x, u, z, u))]       
3.12/1.29	                                                                  
3.12/1.29	
3.12/1.29	The strictly oriented rules are moved into the weak component.
3.12/1.29	
3.12/1.29	We are left with following problem, upon which TcT provides the
3.12/1.29	certificate YES(O(1),O(n^1)).
3.12/1.29	
3.12/1.29	Strict DPs:
3.12/1.29	  { f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u))
3.12/1.29	  , f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u)) }
3.12/1.29	Weak DPs: { perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) }
3.12/1.29	Obligation:
3.12/1.29	  runtime complexity
3.12/1.29	Answer:
3.12/1.29	  YES(O(1),O(n^1))
3.12/1.29	
3.12/1.29	We use the processor 'matrix interpretation of dimension 1' to
3.12/1.29	orient following rules strictly.
3.12/1.29	
3.12/1.29	DPs:
3.12/1.29	  { 1: f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u))
3.12/1.29	  , 2: f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u)) }
3.12/1.29	
3.12/1.29	Sub-proof:
3.12/1.29	----------
3.12/1.29	  The following argument positions are usable:
3.12/1.29	    Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {3}
3.12/1.29	  
3.12/1.29	  TcT has computed the following constructor-based matrix
3.12/1.29	  interpretation satisfying not(EDA).
3.12/1.29	  
3.12/1.29	                      [0] = [0]         
3.12/1.29	                                        
3.12/1.29	                  [s](x1) = [1] x1 + [4]
3.12/1.29	                                        
3.12/1.29	          [minus](x1, x2) = [1] x1 + [4]
3.12/1.29	                                        
3.12/1.29	         [perfectp^#](x1) = [1] x1 + [3]
3.12/1.29	                                        
3.12/1.29	    [f^#](x1, x2, x3, x4) = [1] x1 + [4]
3.12/1.29	                                        
3.12/1.29	                [c_1](x1) = [1] x1 + [3]
3.12/1.29	                                        
3.12/1.29	                [c_2](x1) = [1] x1 + [3]
3.12/1.29	                                        
3.12/1.29	        [c_3](x1, x2, x3) = [1] x3 + [1]
3.12/1.29	  
3.12/1.29	  The order satisfies the following ordering constraints:
3.12/1.29	  
3.12/1.29	         [perfectp^#(s(x))] =  [1] x + [7]                        
3.12/1.29	                            >= [1] x + [7]                        
3.12/1.29	                            =  [c_1(f^#(x, s(0()), s(x), s(x)))]  
3.12/1.29	                                                                  
3.12/1.29	     [f^#(s(x), 0(), z, u)] =  [1] x + [8]                        
3.12/1.29	                            >  [1] x + [7]                        
3.12/1.29	                            =  [c_2(f^#(x, u, minus(z, s(x)), u))]
3.12/1.29	                                                                  
3.12/1.29	    [f^#(s(x), s(y), z, u)] =  [1] x + [8]                        
3.12/1.29	                            >  [1] x + [5]                        
3.12/1.29	                            =  [c_3(x, y, f^#(x, u, z, u))]       
3.12/1.29	                                                                  
3.12/1.29	
3.12/1.29	We return to the main proof. Consider the set of all dependency
3.12/1.29	pairs
3.12/1.29	
3.12/1.29	:
3.12/1.29	  { 1: f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u))
3.12/1.29	  , 2: f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u))
3.12/1.29	  , 3: perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) }
3.12/1.29	
3.12/1.29	Processor 'matrix interpretation of dimension 1' induces the
3.12/1.29	complexity certificate YES(?,O(n^1)) on application of dependency
3.12/1.29	pairs {1,2}. These cover all (indirect) predecessors of dependency
3.12/1.29	pairs {1,2,3}, their number of application is equally bounded. The
3.12/1.29	dependency pairs are shifted into the weak component.
3.12/1.29	
3.12/1.29	We are left with following problem, upon which TcT provides the
3.12/1.29	certificate YES(O(1),O(1)).
3.12/1.29	
3.12/1.29	Weak DPs:
3.12/1.29	  { perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x)))
3.12/1.29	  , f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u))
3.12/1.29	  , f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u)) }
3.12/1.29	Obligation:
3.12/1.29	  runtime complexity
3.12/1.29	Answer:
3.12/1.29	  YES(O(1),O(1))
3.12/1.29	
3.12/1.29	The following weak DPs constitute a sub-graph of the DG that is
3.12/1.29	closed under successors. The DPs are removed.
3.12/1.29	
3.12/1.29	{ perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x)))
3.12/1.29	, f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u))
3.12/1.29	, f^#(s(x), s(y), z, u) -> c_3(x, y, f^#(x, u, z, u)) }
3.12/1.29	
3.12/1.29	We are left with following problem, upon which TcT provides the
3.12/1.29	certificate YES(O(1),O(1)).
3.12/1.29	
3.12/1.29	Rules: Empty
3.12/1.29	Obligation:
3.12/1.29	  runtime complexity
3.12/1.29	Answer:
3.12/1.29	  YES(O(1),O(1))
3.12/1.29	
3.12/1.29	Empty rules are trivially bounded
3.12/1.29	
3.12/1.29	Hurray, we answered YES(O(1),O(n^1))
3.12/1.29	EOF