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: { 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 , U21(tt(), M, N) -> U22(tt(), M, N) , U22(tt(), M, N) -> plus(x(N, M), N) , x(N, s(M)) -> U21(tt(), M, N) , x(N, 0()) -> 0() } 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)) -> U11(tt(), M, N) , U22(tt(), M, N) -> 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. [U11](x1, x2, x3) = x1 + 2*x2 + x3 [tt]() = 1 [U12](x1, x2, x3) = x1 + 2*x2 + x3 [s](x1) = 1 + x1 [plus](x1, x2) = x1 + 2*x2 [U21](x1, x2, x3) = 2*x1 + 2*x1*x2 + 2*x1*x3 + 2*x1^2 + x2 + 2*x2*x3 + 2*x2^2 + 2*x3 [U22](x1, x2, x3) = 2*x1 + 2*x1*x3 + 2*x1^2 + 2*x2 + 2*x2*x3 + 2*x2^2 + 2*x3 [x](x1, x2) = 2*x1 + 2*x1*x2 + 2*x2 + 2*x2^2 [0]() = 0 This order satisfies the following ordering constraints. [U11(tt(), M, N)] = 1 + 2*M + N >= 1 + 2*M + N = [U12(tt(), M, N)] [U12(tt(), M, N)] = 1 + 2*M + N >= 1 + N + 2*M = [s(plus(N, M))] [plus(N, s(M))] = N + 2 + 2*M > 1 + 2*M + N = [U11(tt(), M, N)] [plus(N, 0())] = N >= N = [N] [U21(tt(), M, N)] = 4 + 3*M + 4*N + 2*M*N + 2*M^2 >= 4 + 4*N + 2*M + 2*M*N + 2*M^2 = [U22(tt(), M, N)] [U22(tt(), M, N)] = 4 + 4*N + 2*M + 2*M*N + 2*M^2 > 4*N + 2*N*M + 2*M + 2*M^2 = [plus(x(N, M), N)] [x(N, s(M))] = 4*N + 2*N*M + 4 + 6*M + 2*M^2 >= 4 + 3*M + 4*N + 2*M*N + 2*M^2 = [U21(tt(), M, N)] [x(N, 0())] = 2*N >= = [0()] 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: { U11(tt(), M, N) -> U12(tt(), M, N) , U12(tt(), M, N) -> s(plus(N, M)) , plus(N, 0()) -> N , U21(tt(), M, N) -> U22(tt(), M, N) , x(N, s(M)) -> U21(tt(), M, N) , x(N, 0()) -> 0() } Weak Trs: { plus(N, s(M)) -> U11(tt(), M, N) , U22(tt(), M, N) -> 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, s(M)) -> U21(tt(), 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. [U11](x1, x2, x3) = 2*x1 + 2*x1*x2 + x2 + x3 [tt]() = 1 [U12](x1, x2, x3) = 2*x1 + 3*x2 + x3 [s](x1) = 2 + x1 [plus](x1, x2) = x1 + 3*x2 [U21](x1, x2, x3) = x1 + 2*x1*x3 + 2*x2 + 2*x2*x3 + 2*x3 [U22](x1, x2, x3) = x1 + x1*x2 + 2*x1*x3 + x2 + 2*x2*x3 + 2*x3 [x](x1, x2) = x1 + 2*x1*x2 + 2*x2 [0]() = 0 This order satisfies the following ordering constraints. [U11(tt(), M, N)] = 2 + 3*M + N >= 2 + 3*M + N = [U12(tt(), M, N)] [U12(tt(), M, N)] = 2 + 3*M + N >= 2 + N + 3*M = [s(plus(N, M))] [plus(N, s(M))] = N + 6 + 3*M > 2 + 3*M + N = [U11(tt(), M, N)] [plus(N, 0())] = N >= N = [N] [U21(tt(), M, N)] = 1 + 4*N + 2*M + 2*M*N >= 1 + 2*M + 4*N + 2*M*N = [U22(tt(), M, N)] [U22(tt(), M, N)] = 1 + 2*M + 4*N + 2*M*N > 4*N + 2*N*M + 2*M = [plus(x(N, M), N)] [x(N, s(M))] = 5*N + 2*N*M + 4 + 2*M > 1 + 4*N + 2*M + 2*M*N = [U21(tt(), M, N)] [x(N, 0())] = N >= = [0()] 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: { U11(tt(), M, N) -> U12(tt(), M, N) , U12(tt(), M, N) -> s(plus(N, M)) , plus(N, 0()) -> N , U21(tt(), M, N) -> U22(tt(), M, N) , x(N, 0()) -> 0() } Weak Trs: { plus(N, s(M)) -> U11(tt(), M, N) , U22(tt(), M, N) -> plus(x(N, M), N) , x(N, s(M)) -> U21(tt(), 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 , 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. [U11](x1, x2, x3) = 1 + x1 + 2*x2 + x3 [tt]() = 1 [U12](x1, x2, x3) = 1 + x1 + x1*x2 + x2 + x3 [s](x1) = 2 + x1 [plus](x1, x2) = x1 + 2*x2 [U21](x1, x2, x3) = x1 + x1*x2 + 2*x1*x3 + x2 + x2*x3 + 2*x3 [U22](x1, x2, x3) = x1 + x1*x2 + 2*x1*x3 + x2 + x2*x3 + 2*x3 [x](x1, x2) = 2*x1 + x1*x2 + 2*x2 [0]() = 1 This order satisfies the following ordering constraints. [U11(tt(), M, N)] = 2 + 2*M + N >= 2 + 2*M + N = [U12(tt(), M, N)] [U12(tt(), M, N)] = 2 + 2*M + N >= 2 + N + 2*M = [s(plus(N, M))] [plus(N, s(M))] = N + 4 + 2*M > 2 + 2*M + N = [U11(tt(), M, N)] [plus(N, 0())] = N + 2 > N = [N] [U21(tt(), M, N)] = 1 + 2*M + 4*N + M*N >= 1 + 2*M + 4*N + M*N = [U22(tt(), M, N)] [U22(tt(), M, N)] = 1 + 2*M + 4*N + M*N > 4*N + N*M + 2*M = [plus(x(N, M), N)] [x(N, s(M))] = 4*N + N*M + 4 + 2*M > 1 + 2*M + 4*N + M*N = [U21(tt(), M, N)] [x(N, 0())] = 3*N + 2 > 1 = [0()] 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: { U11(tt(), M, N) -> U12(tt(), M, N) , U12(tt(), M, N) -> s(plus(N, M)) , U21(tt(), M, N) -> U22(tt(), M, N) } Weak Trs: { plus(N, s(M)) -> U11(tt(), M, N) , plus(N, 0()) -> N , U22(tt(), M, N) -> plus(x(N, M), N) , x(N, s(M)) -> U21(tt(), M, N) , x(N, 0()) -> 0() } 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: { U12(tt(), M, N) -> s(plus(N, M)) , U21(tt(), M, N) -> U22(tt(), 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. [U11](x1, x2, x3) = 2*x1 + 2*x2 + x3 [tt]() = 2 [U12](x1, x2, x3) = 2*x1 + 2*x2 + x3 [s](x1) = 2 + x1 [plus](x1, x2) = x1 + 2*x2 [U21](x1, x2, x3) = 1 + 2*x1 + 2*x1*x3 + 2*x2 + 2*x2*x3 + 2*x3 [U22](x1, x2, x3) = 2*x1 + 2*x1*x3 + 2*x2 + 2*x2*x3 + x3 [x](x1, x2) = 1 + 2*x1 + 2*x1*x2 + 2*x2 [0]() = 0 This order satisfies the following ordering constraints. [U11(tt(), M, N)] = 4 + 2*M + N >= 4 + 2*M + N = [U12(tt(), M, N)] [U12(tt(), M, N)] = 4 + 2*M + N > 2 + N + 2*M = [s(plus(N, M))] [plus(N, s(M))] = N + 4 + 2*M >= 4 + 2*M + N = [U11(tt(), M, N)] [plus(N, 0())] = N >= N = [N] [U21(tt(), M, N)] = 5 + 6*N + 2*M + 2*M*N > 4 + 5*N + 2*M + 2*M*N = [U22(tt(), M, N)] [U22(tt(), M, N)] = 4 + 5*N + 2*M + 2*M*N > 1 + 4*N + 2*N*M + 2*M = [plus(x(N, M), N)] [x(N, s(M))] = 5 + 6*N + 2*N*M + 2*M >= 5 + 6*N + 2*M + 2*M*N = [U21(tt(), M, N)] [x(N, 0())] = 1 + 2*N > = [0()] 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: { 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 , U21(tt(), M, N) -> U22(tt(), M, N) , U22(tt(), M, N) -> plus(x(N, M), N) , x(N, s(M)) -> U21(tt(), M, N) , x(N, 0()) -> 0() } 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: { U11(tt(), M, N) -> U12(tt(), 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. [U11](x1, x2, x3) = 3 + x1 + 2*x2 + x3 [tt]() = 1 [U12](x1, x2, x3) = x1 + 2*x1^2 + 2*x2 + x3 [s](x1) = 1 + x1 [plus](x1, x2) = 2 + x1 + 2*x2 [U21](x1, x2, x3) = 1 + x1 + 2*x1*x2 + 2*x1*x3 + 2*x2 + 2*x2*x3 + 2*x2^2 + 2*x3 [U22](x1, x2, x3) = 1 + x1 + 2*x1*x2 + x1*x3 + x2 + 2*x2*x3 + 2*x2^2 + 3*x3 [x](x1, x2) = 2*x1 + 2*x1*x2 + 2*x2 + 2*x2^2 [0]() = 0 This order satisfies the following ordering constraints. [U11(tt(), M, N)] = 4 + 2*M + N > 3 + 2*M + N = [U12(tt(), M, N)] [U12(tt(), M, N)] = 3 + 2*M + N >= 3 + N + 2*M = [s(plus(N, M))] [plus(N, s(M))] = 4 + N + 2*M >= 4 + 2*M + N = [U11(tt(), M, N)] [plus(N, 0())] = 2 + N > N = [N] [U21(tt(), M, N)] = 2 + 4*M + 4*N + 2*M*N + 2*M^2 >= 2 + 3*M + 4*N + 2*M*N + 2*M^2 = [U22(tt(), M, N)] [U22(tt(), M, N)] = 2 + 3*M + 4*N + 2*M*N + 2*M^2 >= 2 + 4*N + 2*N*M + 2*M + 2*M^2 = [plus(x(N, M), N)] [x(N, s(M))] = 4*N + 2*N*M + 4 + 6*M + 2*M^2 > 2 + 4*M + 4*N + 2*M*N + 2*M^2 = [U21(tt(), M, N)] [x(N, 0())] = 2*N >= = [0()] 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 , U21(tt(), M, N) -> U22(tt(), M, N) , U22(tt(), M, N) -> plus(x(N, M), N) , x(N, s(M)) -> U21(tt(), M, N) , x(N, 0()) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^2))