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(), M, N) , U12(tt(), M, N) -> s(plus(N, 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 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 + [3] x2 + [3] x3 + [0] [tt] = [0] [U12](x1, x2, x3) = [3] x1 + [3] x2 + [3] x3 + [0] [s](x1) = [1] x1 + [0] [plus](x1, x2) = [3] x1 + [3] x2 + [0] [0] = [1] This order satisfies the following ordering constraints: [U11(tt(), M, N)] = [3] M + [3] N + [0] >= [3] M + [3] N + [0] = [U12(tt(), M, N)] [U12(tt(), M, N)] = [3] M + [3] N + [0] >= [3] M + [3] N + [0] = [s(plus(N, M))] [plus(N, s(M))] = [3] M + [3] N + [0] >= [3] M + [3] N + [0] = [U11(tt(), M, N)] [plus(N, 0())] = [3] N + [3] > [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(), M, N) , U12(tt(), M, N) -> s(plus(N, M)) , 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 + [3] x3 + [1] [tt] = [0] [U12](x1, x2, x3) = [3] x1 + [2] x2 + [3] x3 + [1] [s](x1) = [1] x1 + [1] [plus](x1, x2) = [3] x1 + [2] x2 + [0] [0] = [3] This order satisfies the following ordering constraints: [U11(tt(), M, N)] = [2] M + [3] N + [1] >= [2] M + [3] N + [1] = [U12(tt(), M, N)] [U12(tt(), M, N)] = [2] M + [3] N + [1] >= [2] M + [3] N + [1] = [s(plus(N, M))] [plus(N, s(M))] = [2] M + [3] N + [2] > [2] M + [3] N + [1] = [U11(tt(), M, N)] [plus(N, 0())] = [3] 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(), M, N) , U12(tt(), M, N) -> s(plus(N, M)) } 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: { U12(tt(), M, N) -> s(plus(N, 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 + [3] x3 + [1] [tt] = [2] [U12](x1, x2, x3) = [2] x1 + [2] x2 + [3] x3 + [1] [s](x1) = [1] x1 + [1] [plus](x1, x2) = [3] x1 + [2] x2 + [3] [0] = [1] This order satisfies the following ordering constraints: [U11(tt(), M, N)] = [2] M + [3] N + [5] >= [2] M + [3] N + [5] = [U12(tt(), M, N)] [U12(tt(), M, N)] = [2] M + [3] N + [5] > [2] M + [3] N + [4] = [s(plus(N, M))] [plus(N, s(M))] = [2] M + [3] N + [5] >= [2] M + [3] N + [5] = [U11(tt(), M, N)] [plus(N, 0())] = [3] N + [5] > [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(), M, N) } Weak Trs: { U12(tt(), M, N) -> s(plus(N, 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 1' to orient following rules strictly. Trs: { U11(tt(), M, N) -> U12(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) = [2] x1 + [2] x2 + [3] x3 + [0] [tt] = [2] [U12](x1, x2, x3) = [1] x1 + [2] x2 + [3] x3 + [0] [s](x1) = [1] x1 + [2] [plus](x1, x2) = [3] x1 + [2] x2 + [0] [0] = [3] This order satisfies the following ordering constraints: [U11(tt(), M, N)] = [2] M + [3] N + [4] > [2] M + [3] N + [2] = [U12(tt(), M, N)] [U12(tt(), M, N)] = [2] M + [3] N + [2] >= [2] M + [3] N + [2] = [s(plus(N, M))] [plus(N, s(M))] = [2] M + [3] N + [4] >= [2] M + [3] N + [4] = [U11(tt(), M, N)] [plus(N, 0())] = [3] 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(1)). Weak Trs: { U11(tt(), M, N) -> U12(tt(), M, N) , U12(tt(), M, N) -> s(plus(N, M)) , 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))