YES(O(1),O(n^2)) 168.07/60.07 YES(O(1),O(n^2)) 168.07/60.07 168.07/60.07 We are left with following problem, upon which TcT provides the 168.07/60.07 certificate YES(O(1),O(n^2)). 168.07/60.07 168.07/60.07 Strict Trs: 168.07/60.07 { minus(minus(x)) -> x 168.07/60.07 , minus(h(x)) -> h(minus(x)) 168.07/60.07 , minus(f(x, y)) -> f(minus(y), minus(x)) } 168.07/60.07 Obligation: 168.07/60.07 derivational complexity 168.07/60.07 Answer: 168.07/60.07 YES(O(1),O(n^2)) 168.07/60.07 168.07/60.07 The weightgap principle applies (using the following nonconstant 168.07/60.07 growth matrix-interpretation) 168.07/60.07 168.07/60.07 TcT has computed the following triangular matrix interpretation. 168.07/60.07 Note that the diagonal of the component-wise maxima of 168.07/60.07 interpretation-entries contains no more than 1 non-zero entries. 168.07/60.07 168.07/60.07 [minus](x1) = [1] x1 + [1] 168.07/60.07 168.07/60.07 [h](x1) = [1] x1 + [2] 168.07/60.07 168.07/60.07 [f](x1, x2) = [1] x1 + [1] x2 + [2] 168.07/60.07 168.07/60.07 The order satisfies the following ordering constraints: 168.07/60.07 168.07/60.07 [minus(minus(x))] = [1] x + [2] 168.07/60.07 > [1] x + [0] 168.07/60.07 = [x] 168.07/60.07 168.07/60.07 [minus(h(x))] = [1] x + [3] 168.07/60.07 >= [1] x + [3] 168.07/60.07 = [h(minus(x))] 168.07/60.07 168.07/60.07 [minus(f(x, y))] = [1] x + [1] y + [3] 168.07/60.07 ? [1] x + [1] y + [4] 168.07/60.07 = [f(minus(y), minus(x))] 168.07/60.07 168.07/60.07 168.07/60.07 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 168.07/60.07 168.07/60.07 We are left with following problem, upon which TcT provides the 168.07/60.07 certificate YES(O(1),O(n^2)). 168.07/60.07 168.07/60.07 Strict Trs: 168.07/60.07 { minus(h(x)) -> h(minus(x)) 168.07/60.07 , minus(f(x, y)) -> f(minus(y), minus(x)) } 168.07/60.07 Weak Trs: { minus(minus(x)) -> x } 168.07/60.07 Obligation: 168.07/60.07 derivational complexity 168.07/60.07 Answer: 168.07/60.07 YES(O(1),O(n^2)) 168.07/60.07 168.07/60.07 We use the processor 'matrix interpretation of dimension 2' to 168.07/60.07 orient following rules strictly. 168.07/60.07 168.07/60.07 Trs: { minus(f(x, y)) -> f(minus(y), minus(x)) } 168.07/60.07 168.07/60.07 The induced complexity on above rules (modulo remaining rules) is 168.07/60.07 YES(?,O(n^2)) . These rules are moved into the corresponding weak 168.07/60.07 component(s). 168.07/60.07 168.07/60.07 Sub-proof: 168.07/60.07 ---------- 168.07/60.07 TcT has computed the following triangular matrix interpretation. 168.07/60.07 168.07/60.07 [minus](x1) = [1 1] x1 + [0] 168.07/60.07 [0 1] [0] 168.07/60.07 168.07/60.07 [h](x1) = [1 0] x1 + [0] 168.07/60.07 [0 1] [0] 168.07/60.07 168.07/60.07 [f](x1, x2) = [1 0] x1 + [1 0] x2 + [2] 168.07/60.07 [0 1] [0 1] [2] 168.07/60.07 168.07/60.07 The order satisfies the following ordering constraints: 168.07/60.07 168.07/60.07 [minus(minus(x))] = [1 2] x + [0] 168.07/60.07 [0 1] [0] 168.07/60.07 >= [1 0] x + [0] 168.07/60.07 [0 1] [0] 168.07/60.07 = [x] 168.07/60.07 168.07/60.07 [minus(h(x))] = [1 1] x + [0] 168.07/60.07 [0 1] [0] 168.07/60.07 >= [1 1] x + [0] 168.07/60.07 [0 1] [0] 168.07/60.07 = [h(minus(x))] 168.07/60.07 168.07/60.07 [minus(f(x, y))] = [1 1] x + [1 1] y + [4] 168.07/60.07 [0 1] [0 1] [2] 168.07/60.07 > [1 1] x + [1 1] y + [2] 168.07/60.07 [0 1] [0 1] [2] 168.07/60.07 = [f(minus(y), minus(x))] 168.07/60.07 168.07/60.07 168.07/60.07 We return to the main proof. 168.07/60.07 168.07/60.07 We are left with following problem, upon which TcT provides the 168.07/60.07 certificate YES(O(1),O(n^2)). 168.07/60.07 168.07/60.07 Strict Trs: { minus(h(x)) -> h(minus(x)) } 168.07/60.07 Weak Trs: 168.07/60.07 { minus(minus(x)) -> x 168.07/60.07 , minus(f(x, y)) -> f(minus(y), minus(x)) } 168.07/60.07 Obligation: 168.07/60.07 derivational complexity 168.07/60.07 Answer: 168.07/60.07 YES(O(1),O(n^2)) 168.07/60.07 168.07/60.07 We use the processor 'matrix interpretation of dimension 2' to 168.07/60.07 orient following rules strictly. 168.07/60.07 168.07/60.07 Trs: { minus(h(x)) -> h(minus(x)) } 168.07/60.07 168.07/60.07 The induced complexity on above rules (modulo remaining rules) is 168.07/60.07 YES(?,O(n^2)) . These rules are moved into the corresponding weak 168.07/60.07 component(s). 168.07/60.07 168.07/60.07 Sub-proof: 168.07/60.07 ---------- 168.07/60.07 TcT has computed the following triangular matrix interpretation. 168.07/60.07 168.07/60.07 [minus](x1) = [1 1] x1 + [0] 168.07/60.07 [0 1] [0] 168.07/60.07 168.07/60.07 [h](x1) = [1 0] x1 + [0] 168.07/60.07 [0 1] [1] 168.07/60.07 168.07/60.07 [f](x1, x2) = [1 0] x1 + [1 0] x2 + [0] 168.07/60.07 [0 1] [0 1] [0] 168.07/60.07 168.07/60.07 The order satisfies the following ordering constraints: 168.07/60.07 168.07/60.07 [minus(minus(x))] = [1 2] x + [0] 168.07/60.07 [0 1] [0] 168.07/60.07 >= [1 0] x + [0] 168.07/60.07 [0 1] [0] 168.07/60.07 = [x] 168.07/60.07 168.07/60.07 [minus(h(x))] = [1 1] x + [1] 168.07/60.07 [0 1] [1] 168.07/60.07 > [1 1] x + [0] 168.07/60.07 [0 1] [1] 168.07/60.07 = [h(minus(x))] 168.07/60.07 168.07/60.07 [minus(f(x, y))] = [1 1] x + [1 1] y + [0] 168.07/60.07 [0 1] [0 1] [0] 168.07/60.07 >= [1 1] x + [1 1] y + [0] 168.07/60.07 [0 1] [0 1] [0] 168.07/60.07 = [f(minus(y), minus(x))] 168.07/60.07 168.07/60.07 168.07/60.07 We return to the main proof. 168.07/60.07 168.07/60.07 We are left with following problem, upon which TcT provides the 168.07/60.07 certificate YES(O(1),O(1)). 168.07/60.07 168.07/60.07 Weak Trs: 168.07/60.07 { minus(minus(x)) -> x 168.07/60.07 , minus(h(x)) -> h(minus(x)) 168.07/60.07 , minus(f(x, y)) -> f(minus(y), minus(x)) } 168.07/60.07 Obligation: 168.07/60.07 derivational complexity 168.07/60.07 Answer: 168.07/60.07 YES(O(1),O(1)) 168.07/60.07 168.07/60.07 Empty rules are trivially bounded 168.07/60.07 168.07/60.07 Hurray, we answered YES(O(1),O(n^2)) 168.07/60.08 EOF