YES(O(1),O(n^1))
0.00/0.50	YES(O(1),O(n^1))
0.00/0.50	
0.00/0.50	We are left with following problem, upon which TcT provides the
0.00/0.50	certificate YES(O(1),O(n^1)).
0.00/0.50	
0.00/0.50	Strict Trs:
0.00/0.50	  { minus(x, 0()) -> x
0.00/0.50	  , minus(s(x), s(y)) -> minus(x, y)
0.00/0.50	  , quot(0(), s(y)) -> 0()
0.00/0.50	  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) }
0.00/0.50	Obligation:
0.00/0.50	  innermost runtime complexity
0.00/0.50	Answer:
0.00/0.50	  YES(O(1),O(n^1))
0.00/0.50	
0.00/0.50	We add the following weak dependency pairs:
0.00/0.50	
0.00/0.50	Strict DPs:
0.00/0.50	  { minus^#(x, 0()) -> c_1()
0.00/0.50	  , minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
0.00/0.50	  , quot^#(0(), s(y)) -> c_3()
0.00/0.50	  , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
0.00/0.50	
0.00/0.50	and mark the set of starting terms.
0.00/0.50	
0.00/0.50	We are left with following problem, upon which TcT provides the
0.00/0.50	certificate YES(O(1),O(n^1)).
0.00/0.50	
0.00/0.50	Strict DPs:
0.00/0.50	  { minus^#(x, 0()) -> c_1()
0.00/0.50	  , minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
0.00/0.50	  , quot^#(0(), s(y)) -> c_3()
0.00/0.50	  , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
0.00/0.50	Strict Trs:
0.00/0.50	  { minus(x, 0()) -> x
0.00/0.50	  , minus(s(x), s(y)) -> minus(x, y)
0.00/0.50	  , quot(0(), s(y)) -> 0()
0.00/0.50	  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) }
0.00/0.50	Obligation:
0.00/0.50	  innermost runtime complexity
0.00/0.50	Answer:
0.00/0.50	  YES(O(1),O(n^1))
0.00/0.50	
0.00/0.50	We replace rewrite rules by usable rules:
0.00/0.50	
0.00/0.50	  Strict Usable Rules:
0.00/0.50	    { minus(x, 0()) -> x
0.00/0.50	    , minus(s(x), s(y)) -> minus(x, y) }
0.00/0.50	
0.00/0.50	We are left with following problem, upon which TcT provides the
0.00/0.50	certificate YES(O(1),O(n^1)).
0.00/0.50	
0.00/0.50	Strict DPs:
0.00/0.50	  { minus^#(x, 0()) -> c_1()
0.00/0.50	  , minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
0.00/0.50	  , quot^#(0(), s(y)) -> c_3()
0.00/0.50	  , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
0.00/0.50	Strict Trs:
0.00/0.50	  { minus(x, 0()) -> x
0.00/0.50	  , minus(s(x), s(y)) -> minus(x, y) }
0.00/0.50	Obligation:
0.00/0.50	  innermost runtime complexity
0.00/0.50	Answer:
0.00/0.50	  YES(O(1),O(n^1))
0.00/0.50	
0.00/0.50	The weightgap principle applies (using the following constant
0.00/0.50	growth matrix-interpretation)
0.00/0.50	
0.00/0.50	The following argument positions are usable:
0.00/0.50	  Uargs(c_2) = {1}, Uargs(quot^#) = {1}, Uargs(c_4) = {1}
0.00/0.50	
0.00/0.50	TcT has computed the following constructor-restricted matrix
0.00/0.50	interpretation.
0.00/0.50	
0.00/0.50	    [minus](x1, x2) = [1 1] x1 + [2]           
0.00/0.50	                      [1 1]      [0]           
0.00/0.50	                                               
0.00/0.50	                [0] = [0]                      
0.00/0.50	                      [0]                      
0.00/0.50	                                               
0.00/0.50	            [s](x1) = [1 1] x1 + [0]           
0.00/0.50	                      [0 0]      [2]           
0.00/0.50	                                               
0.00/0.50	  [minus^#](x1, x2) = [0 0] x1 + [0 0] x2 + [2]
0.00/0.50	                      [1 1]      [1 1]      [2]
0.00/0.50	                                               
0.00/0.50	              [c_1] = [1]                      
0.00/0.50	                      [1]                      
0.00/0.50	                                               
0.00/0.50	          [c_2](x1) = [1 0] x1 + [1]           
0.00/0.50	                      [0 1]      [1]           
0.00/0.50	                                               
0.00/0.50	   [quot^#](x1, x2) = [1 0] x1 + [0]           
0.00/0.50	                      [0 0]      [0]           
0.00/0.50	                                               
0.00/0.50	              [c_3] = [1]                      
0.00/0.50	                      [0]                      
0.00/0.50	                                               
0.00/0.50	          [c_4](x1) = [1 0] x1 + [2]           
0.00/0.50	                      [0 1]      [2]           
0.00/0.50	
0.00/0.50	The order satisfies the following ordering constraints:
0.00/0.50	
0.00/0.50	        [minus(x, 0())] = [1 1] x + [2]                   
0.00/0.50	                          [1 1]     [0]                   
0.00/0.50	                        > [1 0] x + [0]                   
0.00/0.50	                          [0 1]     [0]                   
0.00/0.50	                        = [x]                             
0.00/0.50	                                                          
0.00/0.50	    [minus(s(x), s(y))] = [1 1] x + [4]                   
0.00/0.50	                          [1 1]     [2]                   
0.00/0.50	                        > [1 1] x + [2]                   
0.00/0.50	                          [1 1]     [0]                   
0.00/0.50	                        = [minus(x, y)]                   
0.00/0.50	                                                          
0.00/0.50	      [minus^#(x, 0())] = [0 0] x + [2]                   
0.00/0.50	                          [1 1]     [2]                   
0.00/0.50	                        > [1]                             
0.00/0.50	                          [1]                             
0.00/0.50	                        = [c_1()]                         
0.00/0.50	                                                          
0.00/0.50	  [minus^#(s(x), s(y))] = [0 0] x + [0 0] y + [2]         
0.00/0.50	                          [1 1]     [1 1]     [6]         
0.00/0.50	                        ? [0 0] x + [0 0] y + [3]         
0.00/0.50	                          [1 1]     [1 1]     [3]         
0.00/0.50	                        = [c_2(minus^#(x, y))]            
0.00/0.50	                                                          
0.00/0.50	    [quot^#(0(), s(y))] = [0]                             
0.00/0.50	                          [0]                             
0.00/0.50	                        ? [1]                             
0.00/0.50	                          [0]                             
0.00/0.50	                        = [c_3()]                         
0.00/0.50	                                                          
0.00/0.50	   [quot^#(s(x), s(y))] = [1 1] x + [0]                   
0.00/0.50	                          [0 0]     [0]                   
0.00/0.50	                        ? [1 1] x + [4]                   
0.00/0.50	                          [0 0]     [2]                   
0.00/0.50	                        = [c_4(quot^#(minus(x, y), s(y)))]
0.00/0.50	                                                          
0.00/0.50	
0.00/0.50	Further, it can be verified that all rules not oriented are covered by the weightgap condition.
0.00/0.50	
0.00/0.50	We are left with following problem, upon which TcT provides the
0.00/0.50	certificate YES(O(1),O(n^1)).
0.00/0.50	
0.00/0.50	Strict DPs:
0.00/0.50	  { minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
0.00/0.50	  , quot^#(0(), s(y)) -> c_3()
0.00/0.50	  , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
0.00/0.50	Weak DPs: { minus^#(x, 0()) -> c_1() }
0.00/0.50	Weak Trs:
0.00/0.50	  { minus(x, 0()) -> x
0.00/0.50	  , minus(s(x), s(y)) -> minus(x, y) }
0.00/0.50	Obligation:
0.00/0.50	  innermost runtime complexity
0.00/0.50	Answer:
0.00/0.50	  YES(O(1),O(n^1))
0.00/0.50	
0.00/0.50	We estimate the number of application of {2} by applications of
0.00/0.50	Pre({2}) = {3}. Here rules are labeled as follows:
0.00/0.50	
0.00/0.50	  DPs:
0.00/0.50	    { 1: minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
0.00/0.50	    , 2: quot^#(0(), s(y)) -> c_3()
0.00/0.50	    , 3: quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y)))
0.00/0.50	    , 4: minus^#(x, 0()) -> c_1() }
0.00/0.50	
0.00/0.50	We are left with following problem, upon which TcT provides the
0.00/0.50	certificate YES(O(1),O(n^1)).
0.00/0.50	
0.00/0.50	Strict DPs:
0.00/0.50	  { minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
0.00/0.50	  , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
0.00/0.50	Weak DPs:
0.00/0.50	  { minus^#(x, 0()) -> c_1()
0.00/0.50	  , quot^#(0(), s(y)) -> c_3() }
0.00/0.50	Weak Trs:
0.00/0.50	  { minus(x, 0()) -> x
0.00/0.50	  , minus(s(x), s(y)) -> minus(x, y) }
0.00/0.50	Obligation:
0.00/0.50	  innermost runtime complexity
0.00/0.50	Answer:
0.00/0.50	  YES(O(1),O(n^1))
0.00/0.50	
0.00/0.50	The following weak DPs constitute a sub-graph of the DG that is
0.00/0.50	closed under successors. The DPs are removed.
0.00/0.50	
0.00/0.50	{ minus^#(x, 0()) -> c_1()
0.00/0.50	, quot^#(0(), s(y)) -> c_3() }
0.00/0.50	
0.00/0.50	We are left with following problem, upon which TcT provides the
0.00/0.50	certificate YES(O(1),O(n^1)).
0.00/0.50	
0.00/0.50	Strict DPs:
0.00/0.50	  { minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
0.00/0.50	  , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
0.00/0.50	Weak Trs:
0.00/0.50	  { minus(x, 0()) -> x
0.00/0.50	  , minus(s(x), s(y)) -> minus(x, y) }
0.00/0.50	Obligation:
0.00/0.50	  innermost runtime complexity
0.00/0.50	Answer:
0.00/0.50	  YES(O(1),O(n^1))
0.00/0.50	
0.00/0.50	We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
0.00/0.50	to orient following rules strictly.
0.00/0.50	
0.00/0.50	DPs:
0.00/0.50	  { 1: minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
0.00/0.50	  , 2: quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
0.00/0.50	Trs: { minus(s(x), s(y)) -> minus(x, y) }
0.00/0.50	
0.00/0.50	Sub-proof:
0.00/0.50	----------
0.00/0.50	  The input was oriented with the instance of 'Small Polynomial Path
0.00/0.50	  Order (PS,1-bounded)' as induced by the safe mapping
0.00/0.50	  
0.00/0.50	   safe(minus) = {}, safe(0) = {}, safe(s) = {1}, safe(minus^#) = {2},
0.00/0.50	   safe(c_2) = {}, safe(quot^#) = {2}, safe(c_4) = {}
0.00/0.50	  
0.00/0.50	  and precedence
0.00/0.50	  
0.00/0.50	   minus > minus^#, minus ~ quot^# .
0.00/0.50	  
0.00/0.50	  Following symbols are considered recursive:
0.00/0.50	  
0.00/0.50	   {minus^#, quot^#}
0.00/0.50	  
0.00/0.50	  The recursion depth is 1.
0.00/0.50	  
0.00/0.50	  Further, following argument filtering is employed:
0.00/0.50	  
0.00/0.50	   pi(minus) = 1, pi(0) = [], pi(s) = [1], pi(minus^#) = [1],
0.00/0.50	   pi(c_2) = [1], pi(quot^#) = [1], pi(c_4) = [1]
0.00/0.50	  
0.00/0.50	  Usable defined function symbols are a subset of:
0.00/0.50	  
0.00/0.50	   {minus, minus^#, quot^#}
0.00/0.50	  
0.00/0.50	  For your convenience, here are the satisfied ordering constraints:
0.00/0.50	  
0.00/0.50	    pi(minus^#(s(x), s(y))) =  minus^#(s(; x);)                  
0.00/0.50	                            >  c_2(minus^#(x;);)                 
0.00/0.50	                            =  pi(c_2(minus^#(x, y)))            
0.00/0.50	                                                                 
0.00/0.50	     pi(quot^#(s(x), s(y))) =  quot^#(s(; x);)                   
0.00/0.50	                            >  c_4(quot^#(x;);)                  
0.00/0.50	                            =  pi(c_4(quot^#(minus(x, y), s(y))))
0.00/0.50	                                                                 
0.00/0.50	          pi(minus(x, 0())) =  x                                 
0.00/0.50	                            >= x                                 
0.00/0.50	                            =  pi(x)                             
0.00/0.50	                                                                 
0.00/0.50	      pi(minus(s(x), s(y))) =  s(; x)                            
0.00/0.50	                            >  x                                 
0.00/0.50	                            =  pi(minus(x, y))                   
0.00/0.50	                                                                 
0.00/0.50	
0.00/0.50	The strictly oriented rules are moved into the weak component.
0.00/0.50	
0.00/0.50	We are left with following problem, upon which TcT provides the
0.00/0.50	certificate YES(O(1),O(1)).
0.00/0.50	
0.00/0.50	Weak DPs:
0.00/0.50	  { minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
0.00/0.50	  , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
0.00/0.50	Weak Trs:
0.00/0.50	  { minus(x, 0()) -> x
0.00/0.50	  , minus(s(x), s(y)) -> minus(x, y) }
0.00/0.50	Obligation:
0.00/0.50	  innermost runtime complexity
0.00/0.50	Answer:
0.00/0.50	  YES(O(1),O(1))
0.00/0.50	
0.00/0.50	The following weak DPs constitute a sub-graph of the DG that is
0.00/0.50	closed under successors. The DPs are removed.
0.00/0.50	
0.00/0.50	{ minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
0.00/0.50	, quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
0.00/0.50	
0.00/0.50	We are left with following problem, upon which TcT provides the
0.00/0.50	certificate YES(O(1),O(1)).
0.00/0.50	
0.00/0.50	Weak Trs:
0.00/0.50	  { minus(x, 0()) -> x
0.00/0.50	  , minus(s(x), s(y)) -> minus(x, y) }
0.00/0.50	Obligation:
0.00/0.50	  innermost runtime complexity
0.00/0.50	Answer:
0.00/0.50	  YES(O(1),O(1))
0.00/0.50	
0.00/0.50	No rule is usable, rules are removed from the input problem.
0.00/0.50	
0.00/0.50	We are left with following problem, upon which TcT provides the
0.00/0.50	certificate YES(O(1),O(1)).
0.00/0.50	
0.00/0.50	Rules: Empty
0.00/0.50	Obligation:
0.00/0.50	  innermost runtime complexity
0.00/0.50	Answer:
0.00/0.50	  YES(O(1),O(1))
0.00/0.50	
0.00/0.50	Empty rules are trivially bounded
0.00/0.50	
0.00/0.50	Hurray, we answered YES(O(1),O(n^1))
0.00/0.51	EOF