/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSToCSRProof [EQUIVALENT, 0 ms] (2) CSR (3) CSRRRRProof [EQUIVALENT, 89 ms] (4) CSR (5) CSRRRRProof [EQUIVALENT, 0 ms] (6) CSR (7) CSRRRRProof [EQUIVALENT, 0 ms] (8) CSR (9) RisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) active(nats(N)) -> mark(cons(N, nats(s(N)))) active(zprimes) -> mark(sieve(nats(s(s(0))))) active(filter(X1, X2, X3)) -> filter(active(X1), X2, X3) active(filter(X1, X2, X3)) -> filter(X1, active(X2), X3) active(filter(X1, X2, X3)) -> filter(X1, X2, active(X3)) active(cons(X1, X2)) -> cons(active(X1), X2) active(s(X)) -> s(active(X)) active(sieve(X)) -> sieve(active(X)) active(nats(X)) -> nats(active(X)) filter(mark(X1), X2, X3) -> mark(filter(X1, X2, X3)) filter(X1, mark(X2), X3) -> mark(filter(X1, X2, X3)) filter(X1, X2, mark(X3)) -> mark(filter(X1, X2, X3)) cons(mark(X1), X2) -> mark(cons(X1, X2)) s(mark(X)) -> mark(s(X)) sieve(mark(X)) -> mark(sieve(X)) nats(mark(X)) -> mark(nats(X)) proper(filter(X1, X2, X3)) -> filter(proper(X1), proper(X2), proper(X3)) proper(cons(X1, X2)) -> cons(proper(X1), proper(X2)) proper(0) -> ok(0) proper(s(X)) -> s(proper(X)) proper(sieve(X)) -> sieve(proper(X)) proper(nats(X)) -> nats(proper(X)) proper(zprimes) -> ok(zprimes) filter(ok(X1), ok(X2), ok(X3)) -> ok(filter(X1, X2, X3)) cons(ok(X1), ok(X2)) -> ok(cons(X1, X2)) s(ok(X)) -> ok(s(X)) sieve(ok(X)) -> ok(sieve(X)) nats(ok(X)) -> ok(nats(X)) top(mark(X)) -> top(proper(X)) top(ok(X)) -> top(active(X)) Q is empty. ---------------------------------------- (1) QTRSToCSRProof (EQUIVALENT) The following Q TRS is given: Q restricted rewrite system: The TRS R consists of the following rules: active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) active(nats(N)) -> mark(cons(N, nats(s(N)))) active(zprimes) -> mark(sieve(nats(s(s(0))))) active(filter(X1, X2, X3)) -> filter(active(X1), X2, X3) active(filter(X1, X2, X3)) -> filter(X1, active(X2), X3) active(filter(X1, X2, X3)) -> filter(X1, X2, active(X3)) active(cons(X1, X2)) -> cons(active(X1), X2) active(s(X)) -> s(active(X)) active(sieve(X)) -> sieve(active(X)) active(nats(X)) -> nats(active(X)) filter(mark(X1), X2, X3) -> mark(filter(X1, X2, X3)) filter(X1, mark(X2), X3) -> mark(filter(X1, X2, X3)) filter(X1, X2, mark(X3)) -> mark(filter(X1, X2, X3)) cons(mark(X1), X2) -> mark(cons(X1, X2)) s(mark(X)) -> mark(s(X)) sieve(mark(X)) -> mark(sieve(X)) nats(mark(X)) -> mark(nats(X)) proper(filter(X1, X2, X3)) -> filter(proper(X1), proper(X2), proper(X3)) proper(cons(X1, X2)) -> cons(proper(X1), proper(X2)) proper(0) -> ok(0) proper(s(X)) -> s(proper(X)) proper(sieve(X)) -> sieve(proper(X)) proper(nats(X)) -> nats(proper(X)) proper(zprimes) -> ok(zprimes) filter(ok(X1), ok(X2), ok(X3)) -> ok(filter(X1, X2, X3)) cons(ok(X1), ok(X2)) -> ok(cons(X1, X2)) s(ok(X)) -> ok(s(X)) sieve(ok(X)) -> ok(sieve(X)) nats(ok(X)) -> ok(nats(X)) top(mark(X)) -> top(proper(X)) top(ok(X)) -> top(active(X)) Q is empty. Special symbols used for the transformation (see [GM04]): top: top_1, active: active_1, mark: mark_1, ok: ok_1, proper: proper_1 The replacement map contains the following entries: filter: {1, 2, 3} cons: {1} 0: empty set s: {1} sieve: {1} nats: {1} zprimes: empty set The QTRS contained all rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is complete (and sound). ---------------------------------------- (2) Obligation: Context-sensitive rewrite system: The TRS R consists of the following rules: filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M)) filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M)) sieve(cons(0, Y)) -> cons(0, sieve(Y)) sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N))) nats(N) -> cons(N, nats(s(N))) zprimes -> sieve(nats(s(s(0)))) The replacement map contains the following entries: filter: {1, 2, 3} cons: {1} 0: empty set s: {1} sieve: {1} nats: {1} zprimes: empty set ---------------------------------------- (3) CSRRRRProof (EQUIVALENT) The following CSR is given: Context-sensitive rewrite system: The TRS R consists of the following rules: filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M)) filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M)) sieve(cons(0, Y)) -> cons(0, sieve(Y)) sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N))) nats(N) -> cons(N, nats(s(N))) zprimes -> sieve(nats(s(s(0)))) The replacement map contains the following entries: filter: {1, 2, 3} cons: {1} 0: empty set s: {1} sieve: {1} nats: {1} zprimes: empty set Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(cons(x_1, x_2)) = x_1 POL(filter(x_1, x_2, x_3)) = 2 + x_1 + 2*x_2 + x_3 POL(nats(x_1)) = x_1 POL(s(x_1)) = x_1 POL(sieve(x_1)) = x_1 POL(zprimes) = 1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M)) filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M)) zprimes -> sieve(nats(s(s(0)))) ---------------------------------------- (4) Obligation: Context-sensitive rewrite system: The TRS R consists of the following rules: sieve(cons(0, Y)) -> cons(0, sieve(Y)) sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N))) nats(N) -> cons(N, nats(s(N))) The replacement map contains the following entries: filter: {1, 2, 3} cons: {1} 0: empty set s: {1} sieve: {1} nats: {1} ---------------------------------------- (5) CSRRRRProof (EQUIVALENT) The following CSR is given: Context-sensitive rewrite system: The TRS R consists of the following rules: sieve(cons(0, Y)) -> cons(0, sieve(Y)) sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N))) nats(N) -> cons(N, nats(s(N))) The replacement map contains the following entries: filter: {1, 2, 3} cons: {1} 0: empty set s: {1} sieve: {1} nats: {1} Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(cons(x_1, x_2)) = 1 + x_1 POL(filter(x_1, x_2, x_3)) = x_1 + x_2 + x_3 POL(nats(x_1)) = 1 + x_1 POL(s(x_1)) = x_1 POL(sieve(x_1)) = 1 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: sieve(cons(0, Y)) -> cons(0, sieve(Y)) sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N))) ---------------------------------------- (6) Obligation: Context-sensitive rewrite system: The TRS R consists of the following rules: nats(N) -> cons(N, nats(s(N))) The replacement map contains the following entries: cons: {1} s: {1} nats: {1} ---------------------------------------- (7) CSRRRRProof (EQUIVALENT) The following CSR is given: Context-sensitive rewrite system: The TRS R consists of the following rules: nats(N) -> cons(N, nats(s(N))) The replacement map contains the following entries: cons: {1} s: {1} nats: {1} Used ordering: Polynomial interpretation [POLO]: POL(cons(x_1, x_2)) = x_1 POL(nats(x_1)) = 1 + x_1 POL(s(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: nats(N) -> cons(N, nats(s(N))) ---------------------------------------- (8) Obligation: Context-sensitive rewrite system: R is empty. ---------------------------------------- (9) RisEmptyProof (EQUIVALENT) The CSR R is empty. Hence, termination is trivially proven. ---------------------------------------- (10) YES