/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Outermost Termination of the given OTRS could be proven: (0) OTRS (1) Raffelsieper-Zantema-Transformation [SOUND, 0 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 33 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 0 ms] (6) QTRS (7) AAECC Innermost [EQUIVALENT, 0 ms] (8) QTRS (9) DependencyPairsProof [EQUIVALENT, 0 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QReductionProof [EQUIVALENT, 0 ms] (16) QDP (17) RFCMatchBoundsDPProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Term rewrite system R: The TRS R consists of the following rules: f(x) -> g(f(x)) g(f(x)) -> x g(x) -> a Outermost Strategy. ---------------------------------------- (1) Raffelsieper-Zantema-Transformation (SOUND) We applied the Raffelsieper-Zantema transformation to transform the outermost TRS to a standard TRS. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: down(f(x)) -> up(g(f(x))) down(g(f(x))) -> up(x) down(g(x)) -> up(a) top(up(x)) -> top(down(x)) f_flat(up(x_1)) -> up(f(x_1)) g_flat(up(x_1)) -> up(g(x_1)) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a) = 0 POL(down(x_1)) = 2 + 2*x_1 POL(f(x_1)) = x_1 POL(f_flat(x_1)) = 1 + 2*x_1 POL(g(x_1)) = x_1 POL(g_flat(x_1)) = 2*x_1 POL(top(x_1)) = 2*x_1 POL(up(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: f_flat(up(x_1)) -> up(f(x_1)) g_flat(up(x_1)) -> up(g(x_1)) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: down(f(x)) -> up(g(f(x))) down(g(f(x))) -> up(x) down(g(x)) -> up(a) top(up(x)) -> top(down(x)) Q is empty. ---------------------------------------- (5) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a) = 0 POL(down(x_1)) = 2*x_1 POL(f(x_1)) = 1 + x_1 POL(g(x_1)) = x_1 POL(top(x_1)) = 2*x_1 POL(up(x_1)) = 2*x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: down(g(f(x))) -> up(x) ---------------------------------------- (6) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: down(f(x)) -> up(g(f(x))) down(g(x)) -> up(a) top(up(x)) -> top(down(x)) Q is empty. ---------------------------------------- (7) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is down(f(x)) -> up(g(f(x))) down(g(x)) -> up(a) The TRS R 2 is top(up(x)) -> top(down(x)) The signature Sigma is {top_1} ---------------------------------------- (8) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: down(f(x)) -> up(g(f(x))) down(g(x)) -> up(a) top(up(x)) -> top(down(x)) The set Q consists of the following terms: down(f(x0)) down(g(x0)) top(up(x0)) ---------------------------------------- (9) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(up(x)) -> TOP(down(x)) TOP(up(x)) -> DOWN(x) The TRS R consists of the following rules: down(f(x)) -> up(g(f(x))) down(g(x)) -> up(a) top(up(x)) -> top(down(x)) The set Q consists of the following terms: down(f(x0)) down(g(x0)) top(up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(up(x)) -> TOP(down(x)) The TRS R consists of the following rules: down(f(x)) -> up(g(f(x))) down(g(x)) -> up(a) top(up(x)) -> top(down(x)) The set Q consists of the following terms: down(f(x0)) down(g(x0)) top(up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(up(x)) -> TOP(down(x)) The TRS R consists of the following rules: down(f(x)) -> up(g(f(x))) down(g(x)) -> up(a) The set Q consists of the following terms: down(f(x0)) down(g(x0)) top(up(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. top(up(x0)) ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(up(x)) -> TOP(down(x)) The TRS R consists of the following rules: down(f(x)) -> up(g(f(x))) down(g(x)) -> up(a) The set Q consists of the following terms: down(f(x0)) down(g(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) RFCMatchBoundsDPProof (EQUIVALENT) Finiteness of the DP problem can be shown by a matchbound of 2. As the DP problem is minimal we only have to initialize the certificate graph by the rules of P: TOP(up(x)) -> TOP(down(x)) To find matches we regarded all rules of R and P: down(f(x)) -> up(g(f(x))) down(g(x)) -> up(a) TOP(up(x)) -> TOP(down(x)) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 162, 163, 164, 165, 166, 167, 168, 169 Node 162 is start node and node 163 is final node. Those nodes are connected through the following edges: * 162 to 164 labelled TOP_1(0)* 162 to 167 labelled TOP_1(1)* 162 to 169 labelled TOP_1(2)* 163 to 163 labelled #_1(0)* 164 to 163 labelled down_1(0)* 164 to 165 labelled up_1(1)* 165 to 166 labelled g_1(1)* 165 to 163 labelled a(1)* 166 to 163 labelled f_1(1)* 167 to 165 labelled down_1(1)* 167 to 168 labelled up_1(2)* 168 to 166 labelled a(2)* 169 to 168 labelled down_1(2) ---------------------------------------- (18) YES