/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSToCSRProof [EQUIVALENT, 0 ms] (2) CSR (3) CSRRRRProof [EQUIVALENT, 34 ms] (4) CSR (5) CSDependencyPairsProof [EQUIVALENT, 0 ms] (6) QCSDP (7) QCSDependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(f(X, g(X), Y)) -> mark(f(Y, Y, Y)) active(g(b)) -> mark(c) active(b) -> mark(c) active(g(X)) -> g(active(X)) g(mark(X)) -> mark(g(X)) proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3)) proper(g(X)) -> g(proper(X)) proper(b) -> ok(b) proper(c) -> ok(c) f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3)) g(ok(X)) -> ok(g(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(f(X, g(X), Y)) -> mark(f(Y, Y, Y)) active(g(b)) -> mark(c) active(b) -> mark(c) active(g(X)) -> g(active(X)) g(mark(X)) -> mark(g(X)) proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3)) proper(g(X)) -> g(proper(X)) proper(b) -> ok(b) proper(c) -> ok(c) f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3)) g(ok(X)) -> ok(g(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: f: empty set g: {1} b: empty set c: 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: f(X, g(X), Y) -> f(Y, Y, Y) g(b) -> c b -> c The replacement map contains the following entries: f: empty set g: {1} b: empty set c: empty set ---------------------------------------- (3) CSRRRRProof (EQUIVALENT) The following CSR is given: Context-sensitive rewrite system: The TRS R consists of the following rules: f(X, g(X), Y) -> f(Y, Y, Y) g(b) -> c b -> c The replacement map contains the following entries: f: empty set g: {1} b: empty set c: empty set Used ordering: Polynomial interpretation [POLO]: POL(b) = 2 POL(c) = 1 POL(f(x_1, x_2, x_3)) = 0 POL(g(x_1)) = 2 + 2*x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: g(b) -> c b -> c ---------------------------------------- (4) Obligation: Context-sensitive rewrite system: The TRS R consists of the following rules: f(X, g(X), Y) -> f(Y, Y, Y) The replacement map contains the following entries: f: empty set g: {1} ---------------------------------------- (5) CSDependencyPairsProof (EQUIVALENT) Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem. ---------------------------------------- (6) Obligation: Q-restricted context-sensitive dependency pair problem: The symbols in {g_1} are replacing on all positions. The symbols in {f_3, F_3} are not replacing on any position. The ordinary context-sensitive dependency pairs DP_o are: F(X, g(X), Y) -> F(Y, Y, Y) The TRS R consists of the following rules: f(X, g(X), Y) -> f(Y, Y, Y) Q is empty. ---------------------------------------- (7) QCSDependencyGraphProof (EQUIVALENT) The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 0 SCCs. The rules F(z0, g(z0), z1) -> F(z1, z1, z1) and F(x0, g(x0), x1) -> F(x1, x1, x1) form no chain, because ECap^mu(F(z1, z1, z1)) = F(z1, z1, z1) does not unify with F(x0, g(x0), x1). ---------------------------------------- (8) TRUE