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: { filter(cons(X), 0(), M) -> cons(0()) , filter(cons(X), s(N), M) -> cons(X) , sieve(cons(0())) -> cons(0()) , sieve(cons(s(N))) -> cons(s(N)) , nats(N) -> cons(N) , zprimes() -> sieve(nats(s(s(0())))) } 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: { nats(N) -> cons(N) , zprimes() -> sieve(nats(s(s(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: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [filter](x1, x2, x3) = [3] x1 + [2] x2 + [3] x3 + [0] [cons](x1) = [1] x1 + [0] [0] = [0] [s](x1) = [1] x1 + [0] [sieve](x1) = [1] x1 + [0] [nats](x1) = [2] x1 + [2] [zprimes] = [3] This order satisfies the following ordering constraints: [filter(cons(X), 0(), M)] = [3] X + [3] M + [0] >= [0] = [cons(0())] [filter(cons(X), s(N), M)] = [3] X + [3] M + [2] N + [0] >= [1] X + [0] = [cons(X)] [sieve(cons(0()))] = [0] >= [0] = [cons(0())] [sieve(cons(s(N)))] = [1] N + [0] >= [1] N + [0] = [cons(s(N))] [nats(N)] = [2] N + [2] > [1] N + [0] = [cons(N)] [zprimes()] = [3] > [2] = [sieve(nats(s(s(0()))))] 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: { filter(cons(X), 0(), M) -> cons(0()) , filter(cons(X), s(N), M) -> cons(X) , sieve(cons(0())) -> cons(0()) , sieve(cons(s(N))) -> cons(s(N)) } Weak Trs: { nats(N) -> cons(N) , zprimes() -> sieve(nats(s(s(0())))) } 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: { sieve(cons(0())) -> cons(0()) , sieve(cons(s(N))) -> cons(s(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). [filter](x1, x2, x3) = [3] x1 + [1] x2 + [3] x3 + [0] [cons](x1) = [1] x1 + [0] [0] = [0] [s](x1) = [1] x1 + [0] [sieve](x1) = [2] x1 + [1] [nats](x1) = [2] x1 + [0] [zprimes] = [3] This order satisfies the following ordering constraints: [filter(cons(X), 0(), M)] = [3] X + [3] M + [0] >= [0] = [cons(0())] [filter(cons(X), s(N), M)] = [3] X + [3] M + [1] N + [0] >= [1] X + [0] = [cons(X)] [sieve(cons(0()))] = [1] > [0] = [cons(0())] [sieve(cons(s(N)))] = [2] N + [1] > [1] N + [0] = [cons(s(N))] [nats(N)] = [2] N + [0] >= [1] N + [0] = [cons(N)] [zprimes()] = [3] > [1] = [sieve(nats(s(s(0()))))] 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: { filter(cons(X), 0(), M) -> cons(0()) , filter(cons(X), s(N), M) -> cons(X) } Weak Trs: { sieve(cons(0())) -> cons(0()) , sieve(cons(s(N))) -> cons(s(N)) , nats(N) -> cons(N) , zprimes() -> sieve(nats(s(s(0())))) } 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: { filter(cons(X), s(N), M) -> cons(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). [filter](x1, x2, x3) = [3] x1 + [1] x2 + [3] x3 + [0] [cons](x1) = [1] x1 + [0] [0] = [0] [s](x1) = [1] x1 + [1] [sieve](x1) = [1] x1 + [0] [nats](x1) = [1] x1 + [0] [zprimes] = [3] This order satisfies the following ordering constraints: [filter(cons(X), 0(), M)] = [3] X + [3] M + [0] >= [0] = [cons(0())] [filter(cons(X), s(N), M)] = [3] X + [3] M + [1] N + [1] > [1] X + [0] = [cons(X)] [sieve(cons(0()))] = [0] >= [0] = [cons(0())] [sieve(cons(s(N)))] = [1] N + [1] >= [1] N + [1] = [cons(s(N))] [nats(N)] = [1] N + [0] >= [1] N + [0] = [cons(N)] [zprimes()] = [3] > [2] = [sieve(nats(s(s(0()))))] 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: { filter(cons(X), 0(), M) -> cons(0()) } Weak Trs: { filter(cons(X), s(N), M) -> cons(X) , sieve(cons(0())) -> cons(0()) , sieve(cons(s(N))) -> cons(s(N)) , nats(N) -> cons(N) , zprimes() -> sieve(nats(s(s(0())))) } 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: { filter(cons(X), 0(), M) -> cons(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: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [filter](x1, x2, x3) = [1] x1 + [2] x2 + [3] x3 + [0] [cons](x1) = [1] x1 + [1] [0] = [2] [s](x1) = [1] x1 + [0] [sieve](x1) = [1] x1 + [0] [nats](x1) = [1] x1 + [1] [zprimes] = [3] This order satisfies the following ordering constraints: [filter(cons(X), 0(), M)] = [1] X + [3] M + [5] > [3] = [cons(0())] [filter(cons(X), s(N), M)] = [1] X + [3] M + [2] N + [1] >= [1] X + [1] = [cons(X)] [sieve(cons(0()))] = [3] >= [3] = [cons(0())] [sieve(cons(s(N)))] = [1] N + [1] >= [1] N + [1] = [cons(s(N))] [nats(N)] = [1] N + [1] >= [1] N + [1] = [cons(N)] [zprimes()] = [3] >= [3] = [sieve(nats(s(s(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: { filter(cons(X), 0(), M) -> cons(0()) , filter(cons(X), s(N), M) -> cons(X) , sieve(cons(0())) -> cons(0()) , sieve(cons(s(N))) -> cons(s(N)) , nats(N) -> cons(N) , zprimes() -> sieve(nats(s(s(0())))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))