YES(O(1),O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { U11(tt(), M, N) -> U12(tt(), activate(M), activate(N)) , U12(tt(), M, N) -> s(plus(activate(N), activate(M))) , activate(X) -> X , plus(N, s(M)) -> U11(tt(), M, N) , plus(N, 0()) -> N } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { plus(N, 0()) -> N } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2, x3) = [3] x1 + [2] x2 + [2] x3 + [0] [tt] = [0] [U12](x1, x2, x3) = [3] x1 + [2] x2 + [2] x3 + [0] [activate](x1) = [1] x1 + [0] [s](x1) = [1] x1 + [0] [plus](x1, x2) = [2] x1 + [2] x2 + [0] [0] = [3] This order satisfies the following ordering constraints: [U11(tt(), M, N)] = [2] M + [2] N + [0] >= [2] M + [2] N + [0] = [U12(tt(), activate(M), activate(N))] [U12(tt(), M, N)] = [2] M + [2] N + [0] >= [2] M + [2] N + [0] = [s(plus(activate(N), activate(M)))] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [plus(N, s(M))] = [2] M + [2] N + [0] >= [2] M + [2] N + [0] = [U11(tt(), M, N)] [plus(N, 0())] = [2] N + [6] > [1] N + [0] = [N] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { U11(tt(), M, N) -> U12(tt(), activate(M), activate(N)) , U12(tt(), M, N) -> s(plus(activate(N), activate(M))) , activate(X) -> X , plus(N, s(M)) -> U11(tt(), M, N) } Weak Trs: { plus(N, 0()) -> N } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { plus(N, s(M)) -> U11(tt(), M, N) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2, x3) = [3] x1 + [2] x2 + [1] x3 + [1] [tt] = [0] [U12](x1, x2, x3) = [3] x1 + [2] x2 + [1] x3 + [1] [activate](x1) = [1] x1 + [0] [s](x1) = [1] x1 + [1] [plus](x1, x2) = [1] x1 + [2] x2 + [0] [0] = [3] This order satisfies the following ordering constraints: [U11(tt(), M, N)] = [2] M + [1] N + [1] >= [2] M + [1] N + [1] = [U12(tt(), activate(M), activate(N))] [U12(tt(), M, N)] = [2] M + [1] N + [1] >= [2] M + [1] N + [1] = [s(plus(activate(N), activate(M)))] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [plus(N, s(M))] = [2] M + [1] N + [2] > [2] M + [1] N + [1] = [U11(tt(), M, N)] [plus(N, 0())] = [1] N + [6] > [1] N + [0] = [N] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { U11(tt(), M, N) -> U12(tt(), activate(M), activate(N)) , U12(tt(), M, N) -> s(plus(activate(N), activate(M))) , activate(X) -> X } Weak Trs: { plus(N, s(M)) -> U11(tt(), M, N) , plus(N, 0()) -> N } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { U11(tt(), M, N) -> U12(tt(), activate(M), activate(N)) , U12(tt(), M, N) -> s(plus(activate(N), activate(M))) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2, x3) = [2] x1 + [2] x2 + [1] x3 + [1] [tt] = [2] [U12](x1, x2, x3) = [2] x1 + [2] x2 + [1] x3 + [0] [activate](x1) = [1] x1 + [0] [s](x1) = [1] x1 + [3] [plus](x1, x2) = [1] x1 + [2] x2 + [0] [0] = [3] This order satisfies the following ordering constraints: [U11(tt(), M, N)] = [2] M + [1] N + [5] > [2] M + [1] N + [4] = [U12(tt(), activate(M), activate(N))] [U12(tt(), M, N)] = [2] M + [1] N + [4] > [2] M + [1] N + [3] = [s(plus(activate(N), activate(M)))] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [plus(N, s(M))] = [2] M + [1] N + [6] > [2] M + [1] N + [5] = [U11(tt(), M, N)] [plus(N, 0())] = [1] N + [6] > [1] N + [0] = [N] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { activate(X) -> X } Weak Trs: { U11(tt(), M, N) -> U12(tt(), activate(M), activate(N)) , U12(tt(), M, N) -> s(plus(activate(N), activate(M))) , plus(N, s(M)) -> U11(tt(), M, N) , plus(N, 0()) -> N } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { activate(X) -> X } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA) and not(IDA(1)). [U11](x1, x2, x3) = [1 2] x1 + [1 2] x2 + [2 0] x3 + [0] [0 0] [0 1] [0 2] [3] [tt] = [1] [3] [U12](x1, x2, x3) = [1 0] x1 + [1 2] x2 + [2 0] x3 + [3] [1 0] [0 1] [0 2] [2] [activate](x1) = [1 0] x1 + [1] [0 1] [0] [s](x1) = [1 0] x1 + [1] [0 1] [3] [plus](x1, x2) = [2 0] x1 + [1 2] x2 + [0] [0 2] [0 1] [0] [0] = [0] [0] This order satisfies the following ordering constraints: [U11(tt(), M, N)] = [1 2] M + [2 0] N + [7] [0 1] [0 2] [3] >= [1 2] M + [2 0] N + [7] [0 1] [0 2] [3] = [U12(tt(), activate(M), activate(N))] [U12(tt(), M, N)] = [1 2] M + [2 0] N + [4] [0 1] [0 2] [3] >= [1 2] M + [2 0] N + [4] [0 1] [0 2] [3] = [s(plus(activate(N), activate(M)))] [activate(X)] = [1 0] X + [1] [0 1] [0] > [1 0] X + [0] [0 1] [0] = [X] [plus(N, s(M))] = [1 2] M + [2 0] N + [7] [0 1] [0 2] [3] >= [1 2] M + [2 0] N + [7] [0 1] [0 2] [3] = [U11(tt(), M, N)] [plus(N, 0())] = [2 0] N + [0] [0 2] [0] >= [1 0] N + [0] [0 1] [0] = [N] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { U11(tt(), M, N) -> U12(tt(), activate(M), activate(N)) , U12(tt(), M, N) -> s(plus(activate(N), activate(M))) , activate(X) -> X , plus(N, s(M)) -> U11(tt(), M, N) , plus(N, 0()) -> N } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))