YES(O(1),O(n^2)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { and(tt(), X) -> activate(X) , activate(X) -> X , plus(N, 0()) -> N , plus(N, s(M)) -> s(plus(N, M)) , x(N, 0()) -> 0() , x(N, s(M)) -> plus(x(N, M), N) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'custom shape polynomial interpretation' to orient following rules strictly. Trs: { and(tt(), X) -> activate(X) , activate(X) -> X , x(N, s(M)) -> plus(x(N, M), N) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-restricted polynomial interpretation. [and](x1, x2) = 1 + 3*x1 + 3*x1*x2 + 3*x1^2 + 3*x2 + 3*x2^2 [tt]() = 1 [activate](x1) = 3 + 2*x1 + 3*x1^2 [plus](x1, x2) = x1 + x2 [0]() = 0 [s](x1) = 2 + x1 [x](x1, x2) = 2*x1 + 2*x1*x2 + 2*x2 This order satisfies the following ordering constraints. [and(tt(), X)] = 7 + 6*X + 3*X^2 > 3 + 2*X + 3*X^2 = [activate(X)] [activate(X)] = 3 + 2*X + 3*X^2 > X = [X] [plus(N, 0())] = N >= N = [N] [plus(N, s(M))] = N + 2 + M >= 2 + N + M = [s(plus(N, M))] [x(N, 0())] = 2*N >= = [0()] [x(N, s(M))] = 6*N + 2*N*M + 4 + 2*M > 3*N + 2*N*M + 2*M = [plus(x(N, M), N)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { plus(N, 0()) -> N , plus(N, s(M)) -> s(plus(N, M)) , x(N, 0()) -> 0() } Weak Trs: { and(tt(), X) -> activate(X) , activate(X) -> X , x(N, s(M)) -> plus(x(N, M), N) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'custom shape polynomial interpretation' to orient following rules strictly. Trs: { plus(N, 0()) -> N } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-restricted polynomial interpretation. [and](x1, x2) = 1 + 3*x1 + 3*x1*x2 + 3*x1^2 + 3*x2 + 3*x2^2 [tt]() = 1 [activate](x1) = 3 + x1 + 3*x1^2 [plus](x1, x2) = 3 + x1 + x2 [0]() = 0 [s](x1) = 1 + x1 [x](x1, x2) = x1 + x1*x2 + 2*x2 + 2*x2^2 This order satisfies the following ordering constraints. [and(tt(), X)] = 7 + 6*X + 3*X^2 > 3 + X + 3*X^2 = [activate(X)] [activate(X)] = 3 + X + 3*X^2 > X = [X] [plus(N, 0())] = 3 + N > N = [N] [plus(N, s(M))] = 4 + N + M >= 4 + N + M = [s(plus(N, M))] [x(N, 0())] = N >= = [0()] [x(N, s(M))] = 2*N + N*M + 4 + 6*M + 2*M^2 > 3 + 2*N + N*M + 2*M + 2*M^2 = [plus(x(N, M), N)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { plus(N, s(M)) -> s(plus(N, M)) , x(N, 0()) -> 0() } Weak Trs: { and(tt(), X) -> activate(X) , activate(X) -> X , plus(N, 0()) -> N , x(N, s(M)) -> plus(x(N, M), N) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'custom shape polynomial interpretation' to orient following rules strictly. Trs: { x(N, 0()) -> 0() } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-restricted polynomial interpretation. [and](x1, x2) = 1 + 3*x1 + 3*x1*x2 + 3*x1^2 + 3*x2 + 3*x2^2 [tt]() = 1 [activate](x1) = 3 + 2*x1 + 3*x1^2 [plus](x1, x2) = x1 + x2 [0]() = 0 [s](x1) = 2 + x1 [x](x1, x2) = 2 + 2*x1 + 2*x1*x2 + 2*x2 This order satisfies the following ordering constraints. [and(tt(), X)] = 7 + 6*X + 3*X^2 > 3 + 2*X + 3*X^2 = [activate(X)] [activate(X)] = 3 + 2*X + 3*X^2 > X = [X] [plus(N, 0())] = N >= N = [N] [plus(N, s(M))] = N + 2 + M >= 2 + N + M = [s(plus(N, M))] [x(N, 0())] = 2 + 2*N > = [0()] [x(N, s(M))] = 6 + 6*N + 2*N*M + 2*M > 2 + 3*N + 2*N*M + 2*M = [plus(x(N, M), N)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { plus(N, s(M)) -> s(plus(N, M)) } Weak Trs: { and(tt(), X) -> activate(X) , activate(X) -> X , plus(N, 0()) -> N , x(N, 0()) -> 0() , x(N, s(M)) -> plus(x(N, M), N) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'custom shape polynomial interpretation' to orient following rules strictly. Trs: { plus(N, s(M)) -> s(plus(N, M)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-restricted polynomial interpretation. [and](x1, x2) = 1 + x1 + x1*x2 + x1^2 + 3*x2 + 3*x2^2 [tt]() = 2 [activate](x1) = 3 + x1 + 3*x1^2 [plus](x1, x2) = x1 + 2*x2 [0]() = 0 [s](x1) = 2 + x1 [x](x1, x2) = x1 + 2*x1*x2 + x2 This order satisfies the following ordering constraints. [and(tt(), X)] = 7 + 5*X + 3*X^2 > 3 + X + 3*X^2 = [activate(X)] [activate(X)] = 3 + X + 3*X^2 > X = [X] [plus(N, 0())] = N >= N = [N] [plus(N, s(M))] = N + 4 + 2*M > 2 + N + 2*M = [s(plus(N, M))] [x(N, 0())] = N >= = [0()] [x(N, s(M))] = 5*N + 2*N*M + 2 + M > 3*N + 2*N*M + M = [plus(x(N, M), 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: { and(tt(), X) -> activate(X) , activate(X) -> X , plus(N, 0()) -> N , plus(N, s(M)) -> s(plus(N, M)) , x(N, 0()) -> 0() , x(N, s(M)) -> plus(x(N, M), N) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^2))