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: runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { and(tt(), X) -> activate(X) , 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: ---------- The following argument positions are usable: Uargs(plus) = {1}, Uargs(s) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [and](x1, x2) = [7] x2 + [6] [tt] = [0] [activate](x1) = [1] x1 + [5] [plus](x1, x2) = [1] x1 + [0] [0] = [0] [s](x1) = [1] x1 + [0] [x](x1, x2) = [0] The order satisfies the following ordering constraints: [and(tt(), X)] = [7] X + [6] > [1] X + [5] = [activate(X)] [activate(X)] = [1] X + [5] > [1] X + [0] = [X] [plus(N, 0())] = [1] N + [0] >= [1] N + [0] = [N] [plus(N, s(M))] = [1] N + [0] >= [1] N + [0] = [s(plus(N, M))] [x(N, 0())] = [0] >= [0] = [0()] [x(N, s(M))] = [0] >= [0] = [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() , x(N, s(M)) -> plus(x(N, M), N) } Weak Trs: { and(tt(), X) -> activate(X) , activate(X) -> X } Obligation: runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { x(N, 0()) -> 0() } 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: ---------- The following argument positions are usable: Uargs(plus) = {1}, Uargs(s) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [and](x1, x2) = [1] x1 + [7] x2 + [0] [tt] = [1] [activate](x1) = [1] x1 + [0] [plus](x1, x2) = [1] x1 + [0] [0] = [0] [s](x1) = [1] x1 + [0] [x](x1, x2) = [1] The order satisfies the following ordering constraints: [and(tt(), X)] = [7] X + [1] > [1] X + [0] = [activate(X)] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [plus(N, 0())] = [1] N + [0] >= [1] N + [0] = [N] [plus(N, s(M))] = [1] N + [0] >= [1] N + [0] = [s(plus(N, M))] [x(N, 0())] = [1] > [0] = [0()] [x(N, s(M))] = [1] >= [1] = [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, s(M)) -> plus(x(N, M), N) } Weak Trs: { and(tt(), X) -> activate(X) , activate(X) -> X , x(N, 0()) -> 0() } Obligation: runtime complexity Answer: YES(O(1),O(n^2)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(plus) = {1}, Uargs(s) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [and](x1, x2) = [1] x1 + [1] x2 + [1] [tt] = [0] [activate](x1) = [1] x1 + [0] [plus](x1, x2) = [1] x1 + [4] [0] = [0] [s](x1) = [1] x1 + [4] [x](x1, x2) = [1] x2 + [1] The order satisfies the following ordering constraints: [and(tt(), X)] = [1] X + [1] > [1] X + [0] = [activate(X)] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [plus(N, 0())] = [1] N + [4] > [1] N + [0] = [N] [plus(N, s(M))] = [1] N + [4] ? [1] N + [8] = [s(plus(N, M))] [x(N, 0())] = [1] > [0] = [0()] [x(N, s(M))] = [1] M + [5] >= [1] M + [5] = [plus(x(N, M), N)] Further, it can be verified that all rules not oriented are covered by the weightgap condition. 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, s(M)) -> plus(x(N, M), N) } Weak Trs: { and(tt(), X) -> activate(X) , activate(X) -> X , plus(N, 0()) -> N , x(N, 0()) -> 0() } Obligation: runtime complexity Answer: YES(O(1),O(n^2)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(plus) = {1}, Uargs(s) = {1} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [and](x1, x2) = [1] x1 + [1] x2 + [1] [tt] = [0] [activate](x1) = [1] x1 + [0] [plus](x1, x2) = [1] x1 + [0] [0] = [0] [s](x1) = [1] x1 + [4] [x](x1, x2) = [1] x2 + [0] The order satisfies the following ordering constraints: [and(tt(), X)] = [1] X + [1] > [1] X + [0] = [activate(X)] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [plus(N, 0())] = [1] N + [0] >= [1] N + [0] = [N] [plus(N, s(M))] = [1] N + [0] ? [1] N + [4] = [s(plus(N, M))] [x(N, 0())] = [0] >= [0] = [0()] [x(N, s(M))] = [1] M + [4] > [1] M + [0] = [plus(x(N, M), N)] Further, it can be verified that all rules not oriented are covered by the weightgap condition. 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: 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: ---------- The following argument positions are considered usable: Uargs(plus) = {1}, Uargs(s) = {1} TcT has computed the following constructor-restricted polynomial interpretation. [and](x1, x2) = 2*x2 + 2*x2^2 [tt]() = 0 [activate](x1) = x1 [plus](x1, x2) = 1 + x1 + 2*x2 [0]() = 0 [s](x1) = 2 + x1 [x](x1, x2) = 1 + 2*x1*x2 + 2*x2 This order satisfies the following ordering constraints. [and(tt(), X)] = 2*X + 2*X^2 >= X = [activate(X)] [activate(X)] = X >= X = [X] [plus(N, 0())] = 1 + N > N = [N] [plus(N, s(M))] = 5 + N + 2*M > 3 + N + 2*M = [s(plus(N, M))] [x(N, 0())] = 1 > = [0()] [x(N, s(M))] = 5 + 4*N + 2*N*M + 2*M > 2 + 2*N*M + 2*M + 2*N = [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: runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^2))