3.13/1.60 YES 3.13/1.61 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 3.13/1.61 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.13/1.61 3.13/1.61 3.13/1.61 Termination w.r.t. Q of the given QTRS could be proven: 3.13/1.61 3.13/1.61 (0) QTRS 3.13/1.61 (1) QTRSToCSRProof [SOUND, 0 ms] 3.13/1.61 (2) CSR 3.13/1.61 (3) CSRRRRProof [EQUIVALENT, 60 ms] 3.13/1.61 (4) CSR 3.13/1.61 (5) CSDependencyPairsProof [EQUIVALENT, 0 ms] 3.13/1.61 (6) QCSDP 3.13/1.61 (7) QCSDependencyGraphProof [EQUIVALENT, 0 ms] 3.13/1.61 (8) TRUE 3.13/1.61 3.13/1.61 3.13/1.61 ---------------------------------------- 3.13/1.61 3.13/1.61 (0) 3.13/1.61 Obligation: 3.13/1.61 Q restricted rewrite system: 3.13/1.61 The TRS R consists of the following rules: 3.13/1.61 3.13/1.61 active(f(X, g(X), Y)) -> mark(f(Y, Y, Y)) 3.13/1.61 active(g(b)) -> mark(c) 3.13/1.61 active(b) -> mark(c) 3.13/1.61 active(g(X)) -> g(active(X)) 3.13/1.61 g(mark(X)) -> mark(g(X)) 3.13/1.61 proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3)) 3.13/1.61 proper(g(X)) -> g(proper(X)) 3.13/1.61 proper(b) -> ok(b) 3.13/1.61 proper(c) -> ok(c) 3.13/1.61 f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3)) 3.13/1.61 g(ok(X)) -> ok(g(X)) 3.13/1.61 top(mark(X)) -> top(proper(X)) 3.13/1.61 top(ok(X)) -> top(active(X)) 3.13/1.61 3.13/1.61 The set Q consists of the following terms: 3.13/1.61 3.13/1.61 active(f(x0, g(x0), x1)) 3.13/1.61 active(b) 3.13/1.61 active(g(x0)) 3.13/1.61 g(mark(x0)) 3.13/1.61 proper(f(x0, x1, x2)) 3.13/1.61 proper(g(x0)) 3.13/1.61 proper(b) 3.13/1.61 proper(c) 3.13/1.61 f(ok(x0), ok(x1), ok(x2)) 3.13/1.61 g(ok(x0)) 3.13/1.61 top(mark(x0)) 3.13/1.61 top(ok(x0)) 3.13/1.61 3.13/1.61 3.13/1.61 ---------------------------------------- 3.13/1.61 3.13/1.61 (1) QTRSToCSRProof (SOUND) 3.13/1.61 The following Q TRS is given: Q restricted rewrite system: 3.13/1.61 The TRS R consists of the following rules: 3.13/1.61 3.13/1.61 active(f(X, g(X), Y)) -> mark(f(Y, Y, Y)) 3.13/1.61 active(g(b)) -> mark(c) 3.13/1.61 active(b) -> mark(c) 3.13/1.61 active(g(X)) -> g(active(X)) 3.13/1.61 g(mark(X)) -> mark(g(X)) 3.13/1.61 proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3)) 3.13/1.61 proper(g(X)) -> g(proper(X)) 3.13/1.61 proper(b) -> ok(b) 3.13/1.61 proper(c) -> ok(c) 3.13/1.61 f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3)) 3.13/1.61 g(ok(X)) -> ok(g(X)) 3.13/1.61 top(mark(X)) -> top(proper(X)) 3.13/1.61 top(ok(X)) -> top(active(X)) 3.13/1.61 3.13/1.61 The set Q consists of the following terms: 3.13/1.61 3.13/1.61 active(f(x0, g(x0), x1)) 3.13/1.61 active(b) 3.13/1.61 active(g(x0)) 3.13/1.61 g(mark(x0)) 3.13/1.61 proper(f(x0, x1, x2)) 3.13/1.61 proper(g(x0)) 3.13/1.61 proper(b) 3.13/1.61 proper(c) 3.13/1.61 f(ok(x0), ok(x1), ok(x2)) 3.13/1.61 g(ok(x0)) 3.13/1.61 top(mark(x0)) 3.13/1.61 top(ok(x0)) 3.13/1.61 3.13/1.61 Special symbols used for the transformation (see [GM04]): 3.13/1.61 top: top_1, active: active_1, mark: mark_1, ok: ok_1, proper: proper_1 3.13/1.61 The replacement map contains the following entries: 3.13/1.61 3.13/1.61 f: empty set 3.13/1.61 g: {1} 3.13/1.61 b: empty set 3.13/1.61 c: empty set 3.13/1.61 The QTRS contained just a subset of rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is sound, but not necessarily complete. 3.13/1.61 ---------------------------------------- 3.13/1.61 3.13/1.61 (2) 3.13/1.61 Obligation: 3.13/1.61 Context-sensitive rewrite system: 3.13/1.61 The TRS R consists of the following rules: 3.13/1.61 3.13/1.61 f(X, g(X), Y) -> f(Y, Y, Y) 3.13/1.61 g(b) -> c 3.13/1.61 b -> c 3.13/1.61 3.13/1.61 The replacement map contains the following entries: 3.13/1.61 3.13/1.61 f: empty set 3.13/1.61 g: {1} 3.13/1.61 b: empty set 3.13/1.61 c: empty set 3.13/1.61 3.13/1.61 ---------------------------------------- 3.13/1.61 3.13/1.61 (3) CSRRRRProof (EQUIVALENT) 3.13/1.61 The following CSR is given: Context-sensitive rewrite system: 3.13/1.61 The TRS R consists of the following rules: 3.13/1.61 3.13/1.61 f(X, g(X), Y) -> f(Y, Y, Y) 3.13/1.61 g(b) -> c 3.13/1.61 b -> c 3.13/1.61 3.13/1.61 The replacement map contains the following entries: 3.13/1.61 3.13/1.61 f: empty set 3.13/1.61 g: {1} 3.13/1.61 b: empty set 3.13/1.61 c: empty set 3.13/1.61 Used ordering: 3.13/1.61 Polynomial interpretation [POLO]: 3.13/1.61 3.13/1.61 POL(b) = 2 3.13/1.61 POL(c) = 1 3.13/1.61 POL(f(x_1, x_2, x_3)) = 0 3.13/1.61 POL(g(x_1)) = 2 + 2*x_1 3.13/1.61 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 3.13/1.61 3.13/1.61 g(b) -> c 3.13/1.61 b -> c 3.13/1.61 3.13/1.61 3.13/1.61 3.13/1.61 3.13/1.61 ---------------------------------------- 3.13/1.61 3.13/1.61 (4) 3.13/1.61 Obligation: 3.13/1.61 Context-sensitive rewrite system: 3.13/1.61 The TRS R consists of the following rules: 3.13/1.61 3.13/1.61 f(X, g(X), Y) -> f(Y, Y, Y) 3.13/1.61 3.13/1.61 The replacement map contains the following entries: 3.13/1.61 3.13/1.61 f: empty set 3.13/1.61 g: {1} 3.13/1.61 3.13/1.61 ---------------------------------------- 3.13/1.61 3.13/1.61 (5) CSDependencyPairsProof (EQUIVALENT) 3.13/1.61 Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem. 3.13/1.61 ---------------------------------------- 3.13/1.61 3.13/1.61 (6) 3.13/1.61 Obligation: 3.13/1.61 Q-restricted context-sensitive dependency pair problem: 3.13/1.61 The symbols in {g_1} are replacing on all positions. 3.13/1.61 The symbols in {f_3, F_3} are not replacing on any position. 3.13/1.61 3.13/1.61 The ordinary context-sensitive dependency pairs DP_o are: 3.13/1.61 F(X, g(X), Y) -> F(Y, Y, Y) 3.13/1.61 3.13/1.61 The TRS R consists of the following rules: 3.13/1.61 3.13/1.61 f(X, g(X), Y) -> f(Y, Y, Y) 3.13/1.61 3.13/1.61 Q is empty. 3.13/1.61 3.13/1.61 ---------------------------------------- 3.13/1.61 3.13/1.61 (7) QCSDependencyGraphProof (EQUIVALENT) 3.13/1.61 The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 0 SCCs. 3.13/1.61 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). 3.13/1.61 ---------------------------------------- 3.13/1.61 3.13/1.61 (8) 3.13/1.61 TRUE 3.31/1.64 EOF