YES(O(1),O(1)) 0.00/0.04 YES(O(1),O(1)) 0.00/0.04 0.00/0.04 We are left with following problem, upon which TcT provides the 0.00/0.04 certificate YES(O(1),O(1)). 0.00/0.04 0.00/0.04 Strict Trs: 0.00/0.04 { not(x) -> xor(x, true()) 0.00/0.04 , implies(x, y) -> xor(and(x, y), xor(x, true())) 0.00/0.04 , or(x, y) -> xor(and(x, y), xor(x, y)) 0.00/0.04 , =(x, y) -> xor(x, xor(y, true())) } 0.00/0.04 Obligation: 0.00/0.04 innermost runtime complexity 0.00/0.04 Answer: 0.00/0.04 YES(O(1),O(1)) 0.00/0.04 0.00/0.04 We add the following weak dependency pairs: 0.00/0.04 0.00/0.04 Strict DPs: 0.00/0.04 { not^#(x) -> c_1() 0.00/0.04 , implies^#(x, y) -> c_2() 0.00/0.04 , or^#(x, y) -> c_3() 0.00/0.04 , =^#(x, y) -> c_4() } 0.00/0.04 0.00/0.04 and mark the set of starting terms. 0.00/0.04 0.00/0.04 We are left with following problem, upon which TcT provides the 0.00/0.04 certificate YES(O(1),O(1)). 0.00/0.04 0.00/0.04 Strict DPs: 0.00/0.04 { not^#(x) -> c_1() 0.00/0.04 , implies^#(x, y) -> c_2() 0.00/0.04 , or^#(x, y) -> c_3() 0.00/0.04 , =^#(x, y) -> c_4() } 0.00/0.04 Strict Trs: 0.00/0.04 { not(x) -> xor(x, true()) 0.00/0.04 , implies(x, y) -> xor(and(x, y), xor(x, true())) 0.00/0.04 , or(x, y) -> xor(and(x, y), xor(x, y)) 0.00/0.04 , =(x, y) -> xor(x, xor(y, true())) } 0.00/0.04 Obligation: 0.00/0.04 innermost runtime complexity 0.00/0.04 Answer: 0.00/0.04 YES(O(1),O(1)) 0.00/0.04 0.00/0.04 No rule is usable, rules are removed from the input problem. 0.00/0.04 0.00/0.04 We are left with following problem, upon which TcT provides the 0.00/0.04 certificate YES(O(1),O(1)). 0.00/0.04 0.00/0.04 Strict DPs: 0.00/0.04 { not^#(x) -> c_1() 0.00/0.04 , implies^#(x, y) -> c_2() 0.00/0.04 , or^#(x, y) -> c_3() 0.00/0.04 , =^#(x, y) -> c_4() } 0.00/0.04 Obligation: 0.00/0.04 innermost runtime complexity 0.00/0.04 Answer: 0.00/0.04 YES(O(1),O(1)) 0.00/0.04 0.00/0.04 The weightgap principle applies (using the following constant 0.00/0.04 growth matrix-interpretation) 0.00/0.04 0.00/0.04 The following argument positions are usable: 0.00/0.04 none 0.00/0.04 0.00/0.04 TcT has computed the following constructor-restricted matrix 0.00/0.04 interpretation. 0.00/0.04 0.00/0.04 [not^#](x1) = [2] 0.00/0.04 [1] 0.00/0.04 0.00/0.04 [c_1] = [1] 0.00/0.04 [1] 0.00/0.04 0.00/0.04 [implies^#](x1, x2) = [1] 0.00/0.04 [2] 0.00/0.04 0.00/0.04 [c_2] = [0] 0.00/0.04 [2] 0.00/0.04 0.00/0.04 [or^#](x1, x2) = [1] 0.00/0.04 [1] 0.00/0.04 0.00/0.04 [c_3] = [0] 0.00/0.04 [1] 0.00/0.04 0.00/0.04 [=^#](x1, x2) = [1] 0.00/0.04 [2] 0.00/0.04 0.00/0.04 [c_4] = [0] 0.00/0.04 [1] 0.00/0.04 0.00/0.04 The order satisfies the following ordering constraints: 0.00/0.04 0.00/0.04 [not^#(x)] = [2] 0.00/0.04 [1] 0.00/0.04 > [1] 0.00/0.04 [1] 0.00/0.04 = [c_1()] 0.00/0.04 0.00/0.04 [implies^#(x, y)] = [1] 0.00/0.04 [2] 0.00/0.04 > [0] 0.00/0.04 [2] 0.00/0.04 = [c_2()] 0.00/0.04 0.00/0.04 [or^#(x, y)] = [1] 0.00/0.04 [1] 0.00/0.04 > [0] 0.00/0.04 [1] 0.00/0.04 = [c_3()] 0.00/0.04 0.00/0.04 [=^#(x, y)] = [1] 0.00/0.04 [2] 0.00/0.04 > [0] 0.00/0.04 [1] 0.00/0.04 = [c_4()] 0.00/0.04 0.00/0.04 0.00/0.04 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 0.00/0.04 0.00/0.04 We are left with following problem, upon which TcT provides the 0.00/0.04 certificate YES(O(1),O(1)). 0.00/0.04 0.00/0.04 Weak DPs: 0.00/0.04 { not^#(x) -> c_1() 0.00/0.04 , implies^#(x, y) -> c_2() 0.00/0.04 , or^#(x, y) -> c_3() 0.00/0.04 , =^#(x, y) -> c_4() } 0.00/0.04 Obligation: 0.00/0.04 innermost runtime complexity 0.00/0.04 Answer: 0.00/0.04 YES(O(1),O(1)) 0.00/0.04 0.00/0.04 The following weak DPs constitute a sub-graph of the DG that is 0.00/0.04 closed under successors. The DPs are removed. 0.00/0.04 0.00/0.04 { not^#(x) -> c_1() 0.00/0.04 , implies^#(x, y) -> c_2() 0.00/0.04 , or^#(x, y) -> c_3() 0.00/0.04 , =^#(x, y) -> c_4() } 0.00/0.04 0.00/0.04 We are left with following problem, upon which TcT provides the 0.00/0.04 certificate YES(O(1),O(1)). 0.00/0.04 0.00/0.04 Rules: Empty 0.00/0.04 Obligation: 0.00/0.04 innermost runtime complexity 0.00/0.04 Answer: 0.00/0.04 YES(O(1),O(1)) 0.00/0.04 0.00/0.04 Empty rules are trivially bounded 0.00/0.04 0.00/0.04 Hurray, we answered YES(O(1),O(1)) 0.00/0.04 EOF