YES(O(1),O(n^1))
0.00/0.17	YES(O(1),O(n^1))
0.00/0.17	
0.00/0.17	We are left with following problem, upon which TcT provides the
0.00/0.17	certificate YES(O(1),O(n^1)).
0.00/0.17	
0.00/0.17	Strict Trs:
0.00/0.17	  { and(not(not(x)), y, not(z)) -> and(y, band(x, z), x) }
0.00/0.17	Obligation:
0.00/0.17	  runtime complexity
0.00/0.17	Answer:
0.00/0.17	  YES(O(1),O(n^1))
0.00/0.17	
0.00/0.17	We add the following weak dependency pairs:
0.00/0.17	
0.00/0.17	Strict DPs:
0.00/0.17	  { and^#(not(not(x)), y, not(z)) -> c_1(and^#(y, band(x, z), x)) }
0.00/0.17	
0.00/0.17	and mark the set of starting terms.
0.00/0.17	
0.00/0.17	We are left with following problem, upon which TcT provides the
0.00/0.17	certificate YES(O(1),O(n^1)).
0.00/0.17	
0.00/0.17	Strict DPs:
0.00/0.17	  { and^#(not(not(x)), y, not(z)) -> c_1(and^#(y, band(x, z), x)) }
0.00/0.17	Strict Trs:
0.00/0.17	  { and(not(not(x)), y, not(z)) -> and(y, band(x, z), x) }
0.00/0.17	Obligation:
0.00/0.17	  runtime complexity
0.00/0.17	Answer:
0.00/0.17	  YES(O(1),O(n^1))
0.00/0.17	
0.00/0.17	No rule is usable, rules are removed from the input problem.
0.00/0.17	
0.00/0.17	We are left with following problem, upon which TcT provides the
0.00/0.17	certificate YES(O(1),O(n^1)).
0.00/0.17	
0.00/0.17	Strict DPs:
0.00/0.17	  { and^#(not(not(x)), y, not(z)) -> c_1(and^#(y, band(x, z), x)) }
0.00/0.17	Obligation:
0.00/0.17	  runtime complexity
0.00/0.17	Answer:
0.00/0.17	  YES(O(1),O(n^1))
0.00/0.17	
0.00/0.17	The weightgap principle applies (using the following constant
0.00/0.17	growth matrix-interpretation)
0.00/0.17	
0.00/0.17	The following argument positions are usable:
0.00/0.17	  Uargs(c_1) = {1}
0.00/0.17	
0.00/0.17	TcT has computed the following constructor-restricted matrix
0.00/0.17	interpretation.
0.00/0.17	
0.00/0.17	            [not](x1) = [0]                      
0.00/0.17	                        [2]                      
0.00/0.17	                                                 
0.00/0.17	       [band](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
0.00/0.17	                        [0 0]      [0 0]      [0]
0.00/0.17	                                                 
0.00/0.17	  [and^#](x1, x2, x3) = [0 1] x1 + [0 1] x2 + [0]
0.00/0.17	                        [0 0]      [0 0]      [0]
0.00/0.17	                                                 
0.00/0.17	            [c_1](x1) = [1 0] x1 + [1]           
0.00/0.17	                        [0 1]      [0]           
0.00/0.17	
0.00/0.17	The order satisfies the following ordering constraints:
0.00/0.17	
0.00/0.17	  [and^#(not(not(x)), y, not(z))] = [0 1] y + [2]                 
0.00/0.17	                                    [0 0]     [0]                 
0.00/0.17	                                  > [0 1] y + [1]                 
0.00/0.17	                                    [0 0]     [0]                 
0.00/0.17	                                  = [c_1(and^#(y, band(x, z), x))]
0.00/0.17	                                                                  
0.00/0.17	
0.00/0.17	Further, it can be verified that all rules not oriented are covered by the weightgap condition.
0.00/0.17	
0.00/0.17	We are left with following problem, upon which TcT provides the
0.00/0.17	certificate YES(?,O(n^1)).
0.00/0.17	
0.00/0.17	Weak DPs:
0.00/0.17	  { and^#(not(not(x)), y, not(z)) -> c_1(and^#(y, band(x, z), x)) }
0.00/0.17	Obligation:
0.00/0.17	  runtime complexity
0.00/0.17	Answer:
0.00/0.17	  YES(?,O(n^1))
0.00/0.17	
0.00/0.17	The following weak DPs constitute a sub-graph of the DG that is
0.00/0.17	closed under successors. The DPs are removed.
0.00/0.17	
0.00/0.17	{ and^#(not(not(x)), y, not(z)) -> c_1(and^#(y, band(x, z), x)) }
0.00/0.17	
0.00/0.17	We are left with following problem, upon which TcT provides the
0.00/0.17	certificate YES(?,O(n^1)).
0.00/0.17	
0.00/0.17	Rules: Empty
0.00/0.17	Obligation:
0.00/0.17	  runtime complexity
0.00/0.17	Answer:
0.00/0.17	  YES(?,O(n^1))
0.00/0.17	
0.00/0.17	We employ 'linear path analysis' using the following approximated
0.00/0.17	dependency graph:
0.00/0.17	empty
0.00/0.17	
0.00/0.17	
0.00/0.17	Hurray, we answered YES(O(1),O(n^1))
0.00/0.18	EOF