YES(O(1),O(n^2)) 150.89/60.04 YES(O(1),O(n^2)) 150.89/60.04 150.89/60.04 We are left with following problem, upon which TcT provides the 150.89/60.04 certificate YES(O(1),O(n^2)). 150.89/60.04 150.89/60.04 Strict Trs: 150.89/60.04 { and(tt(), X) -> activate(X) 150.89/60.04 , activate(X) -> X 150.89/60.04 , plus(N, 0()) -> N 150.89/60.04 , plus(N, s(M)) -> s(plus(N, M)) } 150.89/60.04 Obligation: 150.89/60.04 derivational complexity 150.89/60.04 Answer: 150.89/60.04 YES(O(1),O(n^2)) 150.89/60.04 150.89/60.04 The weightgap principle applies (using the following nonconstant 150.89/60.04 growth matrix-interpretation) 150.89/60.04 150.89/60.04 TcT has computed the following triangular matrix interpretation. 150.89/60.04 Note that the diagonal of the component-wise maxima of 150.89/60.04 interpretation-entries contains no more than 1 non-zero entries. 150.89/60.04 150.89/60.04 [and](x1, x2) = [1] x1 + [1] x2 + [2] 150.89/60.04 150.89/60.04 [tt] = [1] 150.89/60.04 150.89/60.04 [activate](x1) = [1] x1 + [2] 150.89/60.04 150.89/60.04 [plus](x1, x2) = [1] x1 + [1] x2 + [1] 150.89/60.04 150.89/60.04 [0] = [1] 150.89/60.04 150.89/60.04 [s](x1) = [1] x1 + [2] 150.89/60.04 150.89/60.04 The order satisfies the following ordering constraints: 150.89/60.04 150.89/60.04 [and(tt(), X)] = [1] X + [3] 150.89/60.04 > [1] X + [2] 150.89/60.04 = [activate(X)] 150.89/60.04 150.89/60.04 [activate(X)] = [1] X + [2] 150.89/60.04 > [1] X + [0] 150.89/60.04 = [X] 150.89/60.04 150.89/60.04 [plus(N, 0())] = [1] N + [2] 150.89/60.04 > [1] N + [0] 150.89/60.04 = [N] 150.89/60.04 150.89/60.04 [plus(N, s(M))] = [1] N + [1] M + [3] 150.89/60.04 >= [1] N + [1] M + [3] 150.89/60.04 = [s(plus(N, M))] 150.89/60.04 150.89/60.04 150.89/60.04 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 150.89/60.04 150.89/60.04 We are left with following problem, upon which TcT provides the 150.89/60.04 certificate YES(O(1),O(n^2)). 150.89/60.04 150.89/60.04 Strict Trs: { plus(N, s(M)) -> s(plus(N, M)) } 150.89/60.04 Weak Trs: 150.89/60.04 { and(tt(), X) -> activate(X) 150.89/60.04 , activate(X) -> X 150.89/60.04 , plus(N, 0()) -> N } 150.89/60.04 Obligation: 150.89/60.04 derivational complexity 150.89/60.04 Answer: 150.89/60.04 YES(O(1),O(n^2)) 150.89/60.04 150.89/60.04 We use the processor 'matrix interpretation of dimension 2' to 150.89/60.04 orient following rules strictly. 150.89/60.04 150.89/60.04 Trs: { plus(N, s(M)) -> s(plus(N, M)) } 150.89/60.04 150.89/60.04 The induced complexity on above rules (modulo remaining rules) is 150.89/60.04 YES(?,O(n^2)) . These rules are moved into the corresponding weak 150.89/60.04 component(s). 150.89/60.04 150.89/60.04 Sub-proof: 150.89/60.04 ---------- 150.89/60.04 TcT has computed the following triangular matrix interpretation. 150.89/60.04 150.89/60.04 [and](x1, x2) = [1 1] x1 + [1 1] x2 + [0] 150.89/60.04 [0 1] [0 1] [1] 150.89/60.04 150.89/60.04 [tt] = [2] 150.89/60.04 [1] 150.89/60.04 150.89/60.04 [activate](x1) = [1 1] x1 + [1] 150.89/60.04 [0 1] [1] 150.89/60.04 150.89/60.04 [plus](x1, x2) = [1 2] x1 + [1 1] x2 + [0] 150.89/60.04 [0 1] [0 1] [0] 150.89/60.04 150.89/60.04 [0] = [1] 150.89/60.04 [2] 150.89/60.04 150.89/60.04 [s](x1) = [1 0] x1 + [0] 150.89/60.04 [0 1] [1] 150.89/60.04 150.89/60.04 The order satisfies the following ordering constraints: 150.89/60.04 150.89/60.04 [and(tt(), X)] = [1 1] X + [3] 150.89/60.04 [0 1] [2] 150.89/60.04 > [1 1] X + [1] 150.89/60.04 [0 1] [1] 150.89/60.04 = [activate(X)] 150.89/60.04 150.89/60.04 [activate(X)] = [1 1] X + [1] 150.89/60.04 [0 1] [1] 150.89/60.04 > [1 0] X + [0] 150.89/60.04 [0 1] [0] 150.89/60.04 = [X] 150.89/60.04 150.89/60.04 [plus(N, 0())] = [1 2] N + [3] 150.89/60.04 [0 1] [2] 150.89/60.04 > [1 0] N + [0] 150.89/60.04 [0 1] [0] 150.89/60.04 = [N] 150.89/60.04 150.89/60.04 [plus(N, s(M))] = [1 2] N + [1 1] M + [1] 150.89/60.04 [0 1] [0 1] [1] 150.89/60.04 > [1 2] N + [1 1] M + [0] 150.89/60.04 [0 1] [0 1] [1] 150.89/60.04 = [s(plus(N, M))] 150.89/60.04 150.89/60.04 150.89/60.04 We return to the main proof. 150.89/60.04 150.89/60.04 We are left with following problem, upon which TcT provides the 150.89/60.04 certificate YES(O(1),O(1)). 150.89/60.04 150.89/60.04 Weak Trs: 150.89/60.04 { and(tt(), X) -> activate(X) 150.89/60.04 , activate(X) -> X 150.89/60.04 , plus(N, 0()) -> N 150.89/60.04 , plus(N, s(M)) -> s(plus(N, M)) } 150.89/60.04 Obligation: 150.89/60.04 derivational complexity 150.89/60.04 Answer: 150.89/60.04 YES(O(1),O(1)) 150.89/60.04 150.89/60.04 Empty rules are trivially bounded 150.89/60.04 150.89/60.04 Hurray, we answered YES(O(1),O(n^2)) 150.89/60.05 EOF