YES(O(1),O(n^2)) 163.03/60.07 YES(O(1),O(n^2)) 163.03/60.07 163.03/60.07 We are left with following problem, upon which TcT provides the 163.03/60.07 certificate YES(O(1),O(n^2)). 163.03/60.07 163.03/60.07 Strict Trs: 163.03/60.07 { +(0(), y) -> y 163.03/60.07 , +(s(x), y) -> s(+(x, y)) 163.03/60.07 , -(x, 0()) -> x 163.03/60.07 , -(0(), y) -> 0() 163.03/60.07 , -(s(x), s(y)) -> -(x, y) } 163.03/60.07 Obligation: 163.03/60.07 derivational complexity 163.03/60.07 Answer: 163.03/60.07 YES(O(1),O(n^2)) 163.03/60.07 163.03/60.07 We use the processor 'matrix interpretation of dimension 1' to 163.03/60.07 orient following rules strictly. 163.03/60.07 163.03/60.07 Trs: 163.03/60.07 { +(0(), y) -> y 163.03/60.07 , -(x, 0()) -> x 163.03/60.07 , -(0(), y) -> 0() 163.03/60.07 , -(s(x), s(y)) -> -(x, y) } 163.03/60.07 163.03/60.07 The induced complexity on above rules (modulo remaining rules) is 163.03/60.07 YES(?,O(n^1)) . These rules are moved into the corresponding weak 163.03/60.07 component(s). 163.03/60.07 163.03/60.07 Sub-proof: 163.03/60.07 ---------- 163.03/60.07 TcT has computed the following triangular matrix interpretation. 163.03/60.07 163.03/60.07 [+](x1, x2) = [1] x1 + [1] x2 + [2] 163.03/60.07 163.03/60.07 [0] = [1] 163.03/60.07 163.03/60.07 [s](x1) = [1] x1 + [2] 163.03/60.07 163.03/60.07 [-](x1, x2) = [1] x1 + [1] x2 + [2] 163.03/60.07 163.03/60.07 The order satisfies the following ordering constraints: 163.03/60.07 163.03/60.07 [+(0(), y)] = [1] y + [3] 163.03/60.07 > [1] y + [0] 163.03/60.07 = [y] 163.03/60.07 163.03/60.07 [+(s(x), y)] = [1] y + [1] x + [4] 163.03/60.07 >= [1] y + [1] x + [4] 163.03/60.07 = [s(+(x, y))] 163.03/60.07 163.03/60.07 [-(x, 0())] = [1] x + [3] 163.03/60.07 > [1] x + [0] 163.03/60.07 = [x] 163.03/60.07 163.03/60.07 [-(0(), y)] = [1] y + [3] 163.03/60.07 > [1] 163.03/60.07 = [0()] 163.03/60.07 163.03/60.07 [-(s(x), s(y))] = [1] y + [1] x + [6] 163.03/60.07 > [1] y + [1] x + [2] 163.03/60.07 = [-(x, y)] 163.03/60.07 163.03/60.07 163.03/60.07 We return to the main proof. 163.03/60.07 163.03/60.07 We are left with following problem, upon which TcT provides the 163.03/60.07 certificate YES(O(1),O(n^2)). 163.03/60.07 163.03/60.07 Strict Trs: { +(s(x), y) -> s(+(x, y)) } 163.03/60.07 Weak Trs: 163.03/60.07 { +(0(), y) -> y 163.03/60.07 , -(x, 0()) -> x 163.03/60.07 , -(0(), y) -> 0() 163.03/60.07 , -(s(x), s(y)) -> -(x, y) } 163.03/60.07 Obligation: 163.03/60.07 derivational complexity 163.03/60.07 Answer: 163.03/60.07 YES(O(1),O(n^2)) 163.03/60.07 163.03/60.07 We use the processor 'matrix interpretation of dimension 2' to 163.03/60.07 orient following rules strictly. 163.03/60.07 163.03/60.07 Trs: { +(s(x), y) -> s(+(x, y)) } 163.03/60.07 163.03/60.07 The induced complexity on above rules (modulo remaining rules) is 163.03/60.07 YES(?,O(n^2)) . These rules are moved into the corresponding weak 163.03/60.07 component(s). 163.03/60.07 163.03/60.07 Sub-proof: 163.03/60.07 ---------- 163.03/60.07 TcT has computed the following triangular matrix interpretation. 163.03/60.07 163.03/60.07 [+](x1, x2) = [1 1] x1 + [1 2] x2 + [0] 163.03/60.07 [0 1] [0 1] [0] 163.03/60.07 163.03/60.07 [0] = [0] 163.03/60.07 [0] 163.03/60.07 163.03/60.07 [s](x1) = [1 0] x1 + [0] 163.03/60.07 [0 1] [1] 163.03/60.07 163.03/60.07 [-](x1, x2) = [1 0] x1 + [1 0] x2 + [0] 163.03/60.07 [0 1] [0 1] [1] 163.03/60.07 163.03/60.07 The order satisfies the following ordering constraints: 163.03/60.07 163.03/60.07 [+(0(), y)] = [1 2] y + [0] 163.03/60.07 [0 1] [0] 163.03/60.07 >= [1 0] y + [0] 163.03/60.07 [0 1] [0] 163.03/60.07 = [y] 163.03/60.07 163.03/60.07 [+(s(x), y)] = [1 2] y + [1 1] x + [1] 163.03/60.07 [0 1] [0 1] [1] 163.03/60.07 > [1 2] y + [1 1] x + [0] 163.03/60.07 [0 1] [0 1] [1] 163.03/60.07 = [s(+(x, y))] 163.03/60.07 163.03/60.07 [-(x, 0())] = [1 0] x + [0] 163.03/60.07 [0 1] [1] 163.03/60.07 >= [1 0] x + [0] 163.03/60.07 [0 1] [0] 163.03/60.07 = [x] 163.03/60.07 163.03/60.07 [-(0(), y)] = [1 0] y + [0] 163.03/60.07 [0 1] [1] 163.03/60.07 >= [0] 163.03/60.07 [0] 163.03/60.07 = [0()] 163.03/60.07 163.03/60.07 [-(s(x), s(y))] = [1 0] y + [1 0] x + [0] 163.03/60.07 [0 1] [0 1] [3] 163.03/60.07 >= [1 0] y + [1 0] x + [0] 163.03/60.07 [0 1] [0 1] [1] 163.03/60.07 = [-(x, y)] 163.03/60.07 163.03/60.07 163.03/60.07 We return to the main proof. 163.03/60.07 163.03/60.07 We are left with following problem, upon which TcT provides the 163.03/60.07 certificate YES(O(1),O(1)). 163.03/60.07 163.03/60.07 Weak Trs: 163.03/60.07 { +(0(), y) -> y 163.03/60.07 , +(s(x), y) -> s(+(x, y)) 163.03/60.07 , -(x, 0()) -> x 163.03/60.07 , -(0(), y) -> 0() 163.03/60.07 , -(s(x), s(y)) -> -(x, y) } 163.03/60.07 Obligation: 163.03/60.07 derivational complexity 163.03/60.07 Answer: 163.03/60.07 YES(O(1),O(1)) 163.03/60.07 163.03/60.07 Empty rules are trivially bounded 163.03/60.07 163.03/60.07 Hurray, we answered YES(O(1),O(n^2)) 163.03/60.08 EOF