YES(O(1),O(n^2)) 157.28/60.02 YES(O(1),O(n^2)) 157.28/60.02 157.28/60.02 We are left with following problem, upon which TcT provides the 157.28/60.02 certificate YES(O(1),O(n^2)). 157.28/60.02 157.28/60.02 Strict Trs: 157.28/60.02 { U11(tt(), M, N) -> U12(tt(), activate(M), activate(N)) 157.28/60.02 , U12(tt(), M, N) -> s(plus(activate(N), activate(M))) 157.28/60.02 , activate(X) -> X 157.28/60.02 , plus(N, s(M)) -> U11(tt(), M, N) 157.28/60.02 , plus(N, 0()) -> N } 157.28/60.02 Obligation: 157.28/60.02 derivational complexity 157.28/60.02 Answer: 157.28/60.02 YES(O(1),O(n^2)) 157.28/60.02 157.28/60.02 We use the processor 'matrix interpretation of dimension 1' to 157.28/60.02 orient following rules strictly. 157.28/60.02 157.28/60.02 Trs: { plus(N, 0()) -> N } 157.28/60.02 157.28/60.02 The induced complexity on above rules (modulo remaining rules) is 157.28/60.02 YES(?,O(n^1)) . These rules are moved into the corresponding weak 157.28/60.02 component(s). 157.28/60.02 157.28/60.02 Sub-proof: 157.28/60.02 ---------- 157.28/60.02 TcT has computed the following triangular matrix interpretation. 157.28/60.02 157.28/60.02 [U11](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0] 157.28/60.02 157.28/60.02 [tt] = [0] 157.28/60.02 157.28/60.02 [U12](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0] 157.28/60.02 157.28/60.02 [activate](x1) = [1] x1 + [0] 157.28/60.02 157.28/60.02 [s](x1) = [1] x1 + [0] 157.28/60.02 157.28/60.02 [plus](x1, x2) = [1] x1 + [1] x2 + [0] 157.28/60.02 157.28/60.02 [0] = [2] 157.28/60.02 157.28/60.02 The order satisfies the following ordering constraints: 157.28/60.02 157.28/60.02 [U11(tt(), M, N)] = [1] M + [1] N + [0] 157.28/60.02 >= [1] M + [1] N + [0] 157.28/60.02 = [U12(tt(), activate(M), activate(N))] 157.28/60.02 157.28/60.02 [U12(tt(), M, N)] = [1] M + [1] N + [0] 157.28/60.02 >= [1] M + [1] N + [0] 157.28/60.02 = [s(plus(activate(N), activate(M)))] 157.28/60.02 157.28/60.02 [activate(X)] = [1] X + [0] 157.28/60.02 >= [1] X + [0] 157.28/60.02 = [X] 157.28/60.02 157.28/60.02 [plus(N, s(M))] = [1] M + [1] N + [0] 157.28/60.02 >= [1] M + [1] N + [0] 157.28/60.02 = [U11(tt(), M, N)] 157.28/60.02 157.28/60.02 [plus(N, 0())] = [1] N + [2] 157.28/60.02 > [1] N + [0] 157.28/60.02 = [N] 157.28/60.02 157.28/60.02 157.28/60.02 We return to the main proof. 157.28/60.02 157.28/60.02 We are left with following problem, upon which TcT provides the 157.28/60.02 certificate YES(O(1),O(n^2)). 157.28/60.02 157.28/60.02 Strict Trs: 157.28/60.02 { U11(tt(), M, N) -> U12(tt(), activate(M), activate(N)) 157.28/60.02 , U12(tt(), M, N) -> s(plus(activate(N), activate(M))) 157.28/60.02 , activate(X) -> X 157.28/60.02 , plus(N, s(M)) -> U11(tt(), M, N) } 157.28/60.02 Weak Trs: { plus(N, 0()) -> N } 157.28/60.02 Obligation: 157.28/60.02 derivational complexity 157.28/60.02 Answer: 157.28/60.02 YES(O(1),O(n^2)) 157.28/60.02 157.28/60.02 The weightgap principle applies (using the following nonconstant 157.28/60.02 growth matrix-interpretation) 157.28/60.02 157.28/60.02 TcT has computed the following triangular matrix interpretation. 157.28/60.02 Note that the diagonal of the component-wise maxima of 157.28/60.02 interpretation-entries contains no more than 1 non-zero entries. 157.28/60.02 157.28/60.02 [U11](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [1] 157.28/60.02 157.28/60.02 [tt] = [0] 157.28/60.02 157.28/60.02 [U12](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0] 157.28/60.02 157.28/60.02 [activate](x1) = [1] x1 + [0] 157.28/60.02 157.28/60.02 [s](x1) = [1] x1 + [0] 157.28/60.02 157.28/60.02 [plus](x1, x2) = [1] x1 + [1] x2 + [0] 157.28/60.02 157.28/60.02 [0] = [2] 157.28/60.02 157.28/60.02 The order satisfies the following ordering constraints: 157.28/60.02 157.28/60.02 [U11(tt(), M, N)] = [1] M + [1] N + [1] 157.28/60.02 > [1] M + [1] N + [0] 157.28/60.02 = [U12(tt(), activate(M), activate(N))] 157.28/60.02 157.28/60.02 [U12(tt(), M, N)] = [1] M + [1] N + [0] 157.28/60.02 >= [1] M + [1] N + [0] 157.28/60.02 = [s(plus(activate(N), activate(M)))] 157.28/60.02 157.28/60.02 [activate(X)] = [1] X + [0] 157.28/60.02 >= [1] X + [0] 157.28/60.02 = [X] 157.28/60.02 157.28/60.02 [plus(N, s(M))] = [1] M + [1] N + [0] 157.28/60.02 ? [1] M + [1] N + [1] 157.28/60.02 = [U11(tt(), M, N)] 157.28/60.02 157.28/60.02 [plus(N, 0())] = [1] N + [2] 157.28/60.02 > [1] N + [0] 157.28/60.02 = [N] 157.28/60.02 157.28/60.02 157.28/60.02 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 157.28/60.02 157.28/60.02 We are left with following problem, upon which TcT provides the 157.28/60.02 certificate YES(O(1),O(n^2)). 157.28/60.02 157.28/60.02 Strict Trs: 157.28/60.02 { U12(tt(), M, N) -> s(plus(activate(N), activate(M))) 157.28/60.02 , activate(X) -> X 157.28/60.02 , plus(N, s(M)) -> U11(tt(), M, N) } 157.28/60.02 Weak Trs: 157.28/60.02 { U11(tt(), M, N) -> U12(tt(), activate(M), activate(N)) 157.28/60.02 , plus(N, 0()) -> N } 157.28/60.02 Obligation: 157.28/60.02 derivational complexity 157.28/60.02 Answer: 157.28/60.02 YES(O(1),O(n^2)) 157.28/60.02 157.28/60.02 The weightgap principle applies (using the following nonconstant 157.28/60.02 growth matrix-interpretation) 157.28/60.02 157.28/60.02 TcT has computed the following triangular matrix interpretation. 157.28/60.02 Note that the diagonal of the component-wise maxima of 157.28/60.02 interpretation-entries contains no more than 1 non-zero entries. 157.28/60.02 157.28/60.02 [U11](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [1] 157.28/60.02 157.28/60.02 [tt] = [0] 157.28/60.02 157.28/60.02 [U12](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [1] 157.28/60.02 157.28/60.02 [activate](x1) = [1] x1 + [0] 157.28/60.02 157.28/60.02 [s](x1) = [1] x1 + [0] 157.28/60.02 157.28/60.02 [plus](x1, x2) = [1] x1 + [1] x2 + [0] 157.28/60.02 157.28/60.02 [0] = [2] 157.28/60.02 157.28/60.02 The order satisfies the following ordering constraints: 157.28/60.02 157.28/60.02 [U11(tt(), M, N)] = [1] M + [1] N + [1] 157.28/60.02 >= [1] M + [1] N + [1] 157.28/60.02 = [U12(tt(), activate(M), activate(N))] 157.28/60.02 157.28/60.02 [U12(tt(), M, N)] = [1] M + [1] N + [1] 157.28/60.02 > [1] M + [1] N + [0] 157.28/60.02 = [s(plus(activate(N), activate(M)))] 157.28/60.02 157.28/60.02 [activate(X)] = [1] X + [0] 157.28/60.02 >= [1] X + [0] 157.28/60.02 = [X] 157.28/60.02 157.28/60.02 [plus(N, s(M))] = [1] M + [1] N + [0] 157.28/60.02 ? [1] M + [1] N + [1] 157.28/60.02 = [U11(tt(), M, N)] 157.28/60.02 157.28/60.02 [plus(N, 0())] = [1] N + [2] 157.28/60.02 > [1] N + [0] 157.28/60.02 = [N] 157.28/60.02 157.28/60.02 157.28/60.02 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 157.28/60.02 157.28/60.02 We are left with following problem, upon which TcT provides the 157.28/60.02 certificate YES(O(1),O(n^2)). 157.28/60.02 157.28/60.02 Strict Trs: 157.28/60.02 { activate(X) -> X 157.28/60.02 , plus(N, s(M)) -> U11(tt(), M, N) } 157.28/60.02 Weak Trs: 157.28/60.02 { U11(tt(), M, N) -> U12(tt(), activate(M), activate(N)) 157.28/60.02 , U12(tt(), M, N) -> s(plus(activate(N), activate(M))) 157.28/60.02 , plus(N, 0()) -> N } 157.28/60.02 Obligation: 157.28/60.02 derivational complexity 157.28/60.02 Answer: 157.28/60.02 YES(O(1),O(n^2)) 157.28/60.02 157.28/60.02 The weightgap principle applies (using the following nonconstant 157.28/60.02 growth matrix-interpretation) 157.28/60.02 157.28/60.02 TcT has computed the following triangular matrix interpretation. 157.28/60.02 Note that the diagonal of the component-wise maxima of 157.28/60.02 interpretation-entries contains no more than 1 non-zero entries. 157.28/60.02 157.28/60.02 [U11](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [2] 157.28/60.02 157.28/60.02 [tt] = [2] 157.28/60.02 157.28/60.02 [U12](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0] 157.28/60.02 157.28/60.02 [activate](x1) = [1] x1 + [1] 157.28/60.02 157.28/60.02 [s](x1) = [1] x1 + [0] 157.28/60.02 157.28/60.02 [plus](x1, x2) = [1] x1 + [1] x2 + [0] 157.28/60.02 157.28/60.02 [0] = [2] 157.28/60.02 157.28/60.02 The order satisfies the following ordering constraints: 157.28/60.02 157.28/60.02 [U11(tt(), M, N)] = [1] M + [1] N + [4] 157.28/60.02 >= [1] M + [1] N + [4] 157.28/60.02 = [U12(tt(), activate(M), activate(N))] 157.28/60.02 157.28/60.02 [U12(tt(), M, N)] = [1] M + [1] N + [2] 157.28/60.02 >= [1] M + [1] N + [2] 157.28/60.02 = [s(plus(activate(N), activate(M)))] 157.28/60.02 157.28/60.02 [activate(X)] = [1] X + [1] 157.28/60.02 > [1] X + [0] 157.28/60.02 = [X] 157.28/60.02 157.28/60.02 [plus(N, s(M))] = [1] M + [1] N + [0] 157.28/60.02 ? [1] M + [1] N + [4] 157.28/60.02 = [U11(tt(), M, N)] 157.28/60.02 157.28/60.02 [plus(N, 0())] = [1] N + [2] 157.28/60.02 > [1] N + [0] 157.28/60.02 = [N] 157.28/60.02 157.28/60.02 157.28/60.02 Further, it can be verified that all rules not oriented are covered by the weightgap condition. 157.28/60.02 157.28/60.02 We are left with following problem, upon which TcT provides the 157.28/60.02 certificate YES(O(1),O(n^2)). 157.28/60.02 157.28/60.02 Strict Trs: { plus(N, s(M)) -> U11(tt(), M, N) } 157.28/60.02 Weak Trs: 157.28/60.02 { U11(tt(), M, N) -> U12(tt(), activate(M), activate(N)) 157.28/60.02 , U12(tt(), M, N) -> s(plus(activate(N), activate(M))) 157.28/60.02 , activate(X) -> X 157.28/60.02 , plus(N, 0()) -> N } 157.28/60.02 Obligation: 157.28/60.02 derivational complexity 157.28/60.02 Answer: 157.28/60.02 YES(O(1),O(n^2)) 157.28/60.02 157.28/60.02 We use the processor 'matrix interpretation of dimension 2' to 157.28/60.02 orient following rules strictly. 157.28/60.02 157.28/60.02 Trs: { plus(N, s(M)) -> U11(tt(), M, N) } 157.28/60.02 157.28/60.02 The induced complexity on above rules (modulo remaining rules) is 157.28/60.02 YES(?,O(n^2)) . These rules are moved into the corresponding weak 157.28/60.02 component(s). 157.28/60.02 157.28/60.02 Sub-proof: 157.28/60.02 ---------- 157.28/60.02 TcT has computed the following triangular matrix interpretation. 157.28/60.02 157.28/60.02 [U11](x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 0] x3 + [0] 157.28/60.02 [0 1] [0 1] [0 1] [1] 157.28/60.02 157.28/60.02 [tt] = [0] 157.28/60.02 [0] 157.28/60.02 157.28/60.02 [U12](x1, x2, x3) = [1 2] x1 + [1 1] x2 + [1 0] x3 + [0] 157.28/60.02 [0 1] [0 1] [0 1] [1] 157.28/60.02 157.28/60.02 [activate](x1) = [1 0] x1 + [0] 157.28/60.02 [0 1] [0] 157.28/60.02 157.28/60.02 [s](x1) = [1 0] x1 + [0] 157.28/60.02 [0 1] [1] 157.28/60.02 157.28/60.02 [plus](x1, x2) = [1 0] x1 + [1 1] x2 + [0] 157.28/60.02 [0 1] [0 1] [0] 157.28/60.02 157.28/60.02 [0] = [1] 157.28/60.02 [2] 157.28/60.02 157.28/60.02 The order satisfies the following ordering constraints: 157.28/60.02 157.28/60.02 [U11(tt(), M, N)] = [1 1] M + [1 0] N + [0] 157.28/60.02 [0 1] [0 1] [1] 157.28/60.02 >= [1 1] M + [1 0] N + [0] 157.28/60.02 [0 1] [0 1] [1] 157.28/60.02 = [U12(tt(), activate(M), activate(N))] 157.28/60.02 157.28/60.02 [U12(tt(), M, N)] = [1 1] M + [1 0] N + [0] 157.28/60.02 [0 1] [0 1] [1] 157.28/60.02 >= [1 1] M + [1 0] N + [0] 157.28/60.02 [0 1] [0 1] [1] 157.28/60.02 = [s(plus(activate(N), activate(M)))] 157.28/60.02 157.28/60.02 [activate(X)] = [1 0] X + [0] 157.28/60.02 [0 1] [0] 157.28/60.02 >= [1 0] X + [0] 157.28/60.02 [0 1] [0] 157.28/60.02 = [X] 157.28/60.02 157.28/60.02 [plus(N, s(M))] = [1 1] M + [1 0] N + [1] 157.28/60.03 [0 1] [0 1] [1] 157.28/60.03 > [1 1] M + [1 0] N + [0] 157.28/60.03 [0 1] [0 1] [1] 157.28/60.03 = [U11(tt(), M, N)] 157.28/60.03 157.28/60.03 [plus(N, 0())] = [1 0] N + [3] 157.28/60.03 [0 1] [2] 157.28/60.03 > [1 0] N + [0] 157.28/60.03 [0 1] [0] 157.28/60.03 = [N] 157.28/60.03 157.28/60.03 157.28/60.03 We return to the main proof. 157.28/60.03 157.28/60.03 We are left with following problem, upon which TcT provides the 157.28/60.03 certificate YES(O(1),O(1)). 157.28/60.03 157.28/60.03 Weak Trs: 157.28/60.03 { U11(tt(), M, N) -> U12(tt(), activate(M), activate(N)) 157.28/60.03 , U12(tt(), M, N) -> s(plus(activate(N), activate(M))) 157.28/60.03 , activate(X) -> X 157.28/60.03 , plus(N, s(M)) -> U11(tt(), M, N) 157.28/60.03 , plus(N, 0()) -> N } 157.28/60.03 Obligation: 157.28/60.03 derivational complexity 157.28/60.03 Answer: 157.28/60.03 YES(O(1),O(1)) 157.28/60.03 157.28/60.03 Empty rules are trivially bounded 157.28/60.03 157.28/60.03 Hurray, we answered YES(O(1),O(n^2)) 157.28/60.04 EOF