/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Outermost Termination of the given OTRS could not be shown: (0) OTRS (1) Thiemann-SpecialC-Transformation [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (13) QDP (14) DependencyGraphProof [EQUIVALENT, 0 ms] (15) TRUE (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) QReductionProof [EQUIVALENT, 2 ms] (20) QDP (21) TransformationProof [EQUIVALENT, 0 ms] (22) QDP (23) TransformationProof [EQUIVALENT, 0 ms] (24) QDP (25) TransformationProof [EQUIVALENT, 0 ms] (26) QDP (27) TransformationProof [EQUIVALENT, 0 ms] (28) QDP (29) TransformationProof [EQUIVALENT, 0 ms] (30) QDP (31) QDPOrderProof [EQUIVALENT, 15 ms] (32) QDP (33) SplitQDPProof [EQUIVALENT, 0 ms] (34) AND (35) QDP (36) SemLabProof [SOUND, 0 ms] (37) QDP (38) QDPOrderProof [EQUIVALENT, 0 ms] (39) QDP (40) PisEmptyProof [SOUND, 0 ms] (41) TRUE (42) QDP (43) UsableRulesProof [EQUIVALENT, 0 ms] (44) QDP (45) QReductionProof [EQUIVALENT, 0 ms] (46) QDP (47) Trivial-Transformation [SOUND, 0 ms] (48) QTRS (49) DependencyPairsProof [EQUIVALENT, 0 ms] (50) QDP (51) MNOCProof [EQUIVALENT, 0 ms] (52) QDP (53) TransformationProof [EQUIVALENT, 0 ms] (54) QDP (55) MNOCProof [EQUIVALENT, 0 ms] (56) QDP (57) NonTerminationLoopProof [COMPLETE, 0 ms] (58) NO ---------------------------------------- (0) Obligation: Term rewrite system R: The TRS R consists of the following rules: g(x, x) -> f(f(x, x), x) f(x, x) -> g(x, x) f(x, y) -> y Outermost Strategy. ---------------------------------------- (1) Thiemann-SpecialC-Transformation (EQUIVALENT) We applied the Thiemann-SpecialC transformation to transform the outermost TRS to an innermost TRS. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: top(go_up(x)) -> top(reduce(x)) reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_g(x, x) -> result_g(f(f(x, x), x)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_g(result_g(x)) -> go_up(x) check_f(result_f(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2) check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2)) in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2)) The set Q consists of the following terms: top(go_up(x0)) reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) TOP(go_up(x)) -> REDUCE(x) REDUCE(g(x_1, x_2)) -> CHECK_G(redex_g(x_1, x_2)) REDUCE(g(x_1, x_2)) -> REDEX_G(x_1, x_2) REDUCE(f(x_1, x_2)) -> CHECK_F(redex_f(x_1, x_2)) REDUCE(f(x_1, x_2)) -> REDEX_F(x_1, x_2) CHECK_G(redex_g(x_1, x_2)) -> IN_G_1(reduce(x_1), x_2) CHECK_G(redex_g(x_1, x_2)) -> REDUCE(x_1) CHECK_G(redex_g(x_1, x_2)) -> IN_G_2(x_1, reduce(x_2)) CHECK_G(redex_g(x_1, x_2)) -> REDUCE(x_2) CHECK_F(redex_f(x_1, x_2)) -> IN_F_1(reduce(x_1), x_2) CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_1) CHECK_F(redex_f(x_1, x_2)) -> IN_F_2(x_1, reduce(x_2)) CHECK_F(redex_f(x_1, x_2)) -> REDUCE(x_2) The TRS R consists of the following rules: top(go_up(x)) -> top(reduce(x)) reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_g(x, x) -> result_g(f(f(x, x), x)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_g(result_g(x)) -> go_up(x) check_f(result_f(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2) check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2)) in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2)) The set Q consists of the following terms: top(go_up(x0)) reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 10 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: CHECK_G(redex_g(x_1, x_2)) -> REDUCE(x_1) REDUCE(g(x_1, x_2)) -> CHECK_G(redex_g(x_1, x_2)) CHECK_G(redex_g(x_1, x_2)) -> REDUCE(x_2) The TRS R consists of the following rules: top(go_up(x)) -> top(reduce(x)) reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_g(x, x) -> result_g(f(f(x, x), x)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_g(result_g(x)) -> go_up(x) check_f(result_f(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2) check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2)) in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2)) The set Q consists of the following terms: top(go_up(x0)) reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) 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. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: CHECK_G(redex_g(x_1, x_2)) -> REDUCE(x_1) REDUCE(g(x_1, x_2)) -> CHECK_G(redex_g(x_1, x_2)) CHECK_G(redex_g(x_1, x_2)) -> REDUCE(x_2) The TRS R consists of the following rules: redex_g(x, x) -> result_g(f(f(x, x), x)) The set Q consists of the following terms: top(go_up(x0)) reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) 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(go_up(x0)) reduce(g(x0, x1)) reduce(f(x0, x1)) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: CHECK_G(redex_g(x_1, x_2)) -> REDUCE(x_1) REDUCE(g(x_1, x_2)) -> CHECK_G(redex_g(x_1, x_2)) CHECK_G(redex_g(x_1, x_2)) -> REDUCE(x_2) The TRS R consists of the following rules: redex_g(x, x) -> result_g(f(f(x, x), x)) The set Q consists of the following terms: redex_g(x0, x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) UsableRulesReductionPairsProof (EQUIVALENT) By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well. The following dependency pairs can be deleted: REDUCE(g(x_1, x_2)) -> CHECK_G(redex_g(x_1, x_2)) No rules are removed from R. Used ordering: POLO with Polynomial interpretation [POLO]: POL(CHECK_G(x_1)) = 2*x_1 POL(REDUCE(x_1)) = 2*x_1 POL(f(x_1, x_2)) = x_1 + x_2 POL(g(x_1, x_2)) = 2*x_1 + 2*x_2 POL(redex_g(x_1, x_2)) = 2*x_1 + x_2 POL(result_g(x_1)) = x_1 ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: CHECK_G(redex_g(x_1, x_2)) -> REDUCE(x_1) CHECK_G(redex_g(x_1, x_2)) -> REDUCE(x_2) The TRS R consists of the following rules: redex_g(x, x) -> result_g(f(f(x, x), x)) The set Q consists of the following terms: redex_g(x0, x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes. ---------------------------------------- (15) TRUE ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) The TRS R consists of the following rules: top(go_up(x)) -> top(reduce(x)) reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_g(x, x) -> result_g(f(f(x, x), x)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_g(result_g(x)) -> go_up(x) check_f(result_f(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) check_f(redex_f(x_1, x_2)) -> in_f_1(reduce(x_1), x_2) check_f(redex_f(x_1, x_2)) -> in_f_2(x_1, reduce(x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_f_1(go_up(x_1), x_2) -> go_up(f(x_1, x_2)) in_f_2(x_1, go_up(x_2)) -> go_up(f(x_1, x_2)) The set Q consists of the following terms: top(go_up(x0)) reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) 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. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) The TRS R consists of the following rules: reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_f(result_f(x)) -> go_up(x) redex_g(x, x) -> result_g(f(f(x, x), x)) check_g(result_g(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) The set Q consists of the following terms: top(go_up(x0)) reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) 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(go_up(x0)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) The TRS R consists of the following rules: reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_f(result_f(x)) -> go_up(x) redex_g(x, x) -> result_g(f(f(x, x), x)) check_g(result_g(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) The set Q consists of the following terms: reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule TOP(go_up(x)) -> TOP(reduce(x)) at position [0] we obtained the following new rules [LPAR04]: (TOP(go_up(g(x0, x1))) -> TOP(check_g(redex_g(x0, x1))),TOP(go_up(g(x0, x1))) -> TOP(check_g(redex_g(x0, x1)))) (TOP(go_up(f(x0, x1))) -> TOP(check_f(redex_f(x0, x1))),TOP(go_up(f(x0, x1))) -> TOP(check_f(redex_f(x0, x1)))) ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(g(x0, x1))) -> TOP(check_g(redex_g(x0, x1))) TOP(go_up(f(x0, x1))) -> TOP(check_f(redex_f(x0, x1))) The TRS R consists of the following rules: reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_f(result_f(x)) -> go_up(x) redex_g(x, x) -> result_g(f(f(x, x), x)) check_g(result_g(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) The set Q consists of the following terms: reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule TOP(go_up(f(x0, x1))) -> TOP(check_f(redex_f(x0, x1))) at position [0,0] we obtained the following new rules [LPAR04]: (TOP(go_up(f(x0, x0))) -> TOP(check_f(result_f(g(x0, x0)))),TOP(go_up(f(x0, x0))) -> TOP(check_f(result_f(g(x0, x0))))) (TOP(go_up(f(x0, x1))) -> TOP(check_f(result_f(x1))),TOP(go_up(f(x0, x1))) -> TOP(check_f(result_f(x1)))) ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(g(x0, x1))) -> TOP(check_g(redex_g(x0, x1))) TOP(go_up(f(x0, x0))) -> TOP(check_f(result_f(g(x0, x0)))) TOP(go_up(f(x0, x1))) -> TOP(check_f(result_f(x1))) The TRS R consists of the following rules: reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_f(result_f(x)) -> go_up(x) redex_g(x, x) -> result_g(f(f(x, x), x)) check_g(result_g(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) The set Q consists of the following terms: reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule TOP(go_up(f(x0, x0))) -> TOP(check_f(result_f(g(x0, x0)))) at position [0] we obtained the following new rules [LPAR04]: (TOP(go_up(f(x0, x0))) -> TOP(go_up(g(x0, x0))),TOP(go_up(f(x0, x0))) -> TOP(go_up(g(x0, x0)))) ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(g(x0, x1))) -> TOP(check_g(redex_g(x0, x1))) TOP(go_up(f(x0, x1))) -> TOP(check_f(result_f(x1))) TOP(go_up(f(x0, x0))) -> TOP(go_up(g(x0, x0))) The TRS R consists of the following rules: reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_f(result_f(x)) -> go_up(x) redex_g(x, x) -> result_g(f(f(x, x), x)) check_g(result_g(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) The set Q consists of the following terms: reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule TOP(go_up(f(x0, x1))) -> TOP(check_f(result_f(x1))) at position [0] we obtained the following new rules [LPAR04]: (TOP(go_up(f(x0, x1))) -> TOP(go_up(x1)),TOP(go_up(f(x0, x1))) -> TOP(go_up(x1))) ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(g(x0, x1))) -> TOP(check_g(redex_g(x0, x1))) TOP(go_up(f(x0, x0))) -> TOP(go_up(g(x0, x0))) TOP(go_up(f(x0, x1))) -> TOP(go_up(x1)) The TRS R consists of the following rules: reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_f(result_f(x)) -> go_up(x) redex_g(x, x) -> result_g(f(f(x, x), x)) check_g(result_g(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) The set Q consists of the following terms: reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule TOP(go_up(f(x0, x1))) -> TOP(go_up(x1)) we obtained the following new rules [LPAR04]: (TOP(go_up(f(x0, g(y_0, y_1)))) -> TOP(go_up(g(y_0, y_1))),TOP(go_up(f(x0, g(y_0, y_1)))) -> TOP(go_up(g(y_0, y_1)))) (TOP(go_up(f(x0, f(y_0, y_1)))) -> TOP(go_up(f(y_0, y_1))),TOP(go_up(f(x0, f(y_0, y_1)))) -> TOP(go_up(f(y_0, y_1)))) ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(g(x0, x1))) -> TOP(check_g(redex_g(x0, x1))) TOP(go_up(f(x0, x0))) -> TOP(go_up(g(x0, x0))) TOP(go_up(f(x0, g(y_0, y_1)))) -> TOP(go_up(g(y_0, y_1))) TOP(go_up(f(x0, f(y_0, y_1)))) -> TOP(go_up(f(y_0, y_1))) The TRS R consists of the following rules: reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_f(result_f(x)) -> go_up(x) redex_g(x, x) -> result_g(f(f(x, x), x)) check_g(result_g(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) The set Q consists of the following terms: reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. TOP(go_up(f(x0, g(y_0, y_1)))) -> TOP(go_up(g(y_0, y_1))) TOP(go_up(f(x0, f(y_0, y_1)))) -> TOP(go_up(f(y_0, y_1))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(TOP(x_1)) = x_1 POL(check_f(x_1)) = x_1 POL(check_g(x_1)) = x_1 POL(f(x_1, x_2)) = 1 + x_2 POL(g(x_1, x_2)) = 1 + x_1 POL(go_up(x_1)) = x_1 POL(in_g_1(x_1, x_2)) = 1 + x_1 POL(in_g_2(x_1, x_2)) = 1 + x_1 POL(redex_f(x_1, x_2)) = 1 + x_2 POL(redex_g(x_1, x_2)) = 1 + x_1 POL(reduce(x_1)) = x_1 POL(result_f(x_1)) = x_1 POL(result_g(x_1)) = x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: redex_g(x, x) -> result_g(f(f(x, x), x)) check_g(result_g(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_f(result_f(x)) -> go_up(x) ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(g(x0, x1))) -> TOP(check_g(redex_g(x0, x1))) TOP(go_up(f(x0, x0))) -> TOP(go_up(g(x0, x0))) The TRS R consists of the following rules: reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_f(result_f(x)) -> go_up(x) redex_g(x, x) -> result_g(f(f(x, x), x)) check_g(result_g(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) The set Q consists of the following terms: reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) SplitQDPProof (EQUIVALENT) We show in the first subproof that some pairs and rules can be removed, afterwards, we continue with the remaining DP-Problem ---------------------------------------- (34) Complex Obligation (AND) ---------------------------------------- (35) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(g(x0, x1))) -> TOP(check_g(redex_g(x0, x1))) TOP(go_up(f(x0, x0))) -> TOP(go_up(g(x0, x0))) The TRS R consists of the following rules: reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_f(result_f(x)) -> go_up(x) redex_g(x, x) -> result_g(f(f(x, x), x)) check_g(result_g(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) The set Q consists of the following terms: reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (36) SemLabProof (SOUND) We found the following model for the rules of the TRSs R and P. Interpretation over the domain with elements from 0 to 1. result_f: 0 reduce: 0 in_g_1: 0 redex_f: 0 check_f: 0 in_g_2: 0 TOP: 0 g: 0 go_up: 0 result_g: 0 f: 1 + x0 check_g: 0 redex_g: 0 By semantic labelling [SEMLAB] we obtain the following labelled QDP problem. ---------------------------------------- (37) Obligation: Q DP problem: The TRS P consists of the following rules: TOP.0(go_up.0(g.0-0(x0, x1))) -> TOP.0(check_g.0(redex_g.0-0(x0, x1))) TOP.0(go_up.0(g.0-1(x0, x1))) -> TOP.0(check_g.0(redex_g.0-1(x0, x1))) TOP.0(go_up.0(g.1-0(x0, x1))) -> TOP.0(check_g.0(redex_g.1-0(x0, x1))) TOP.0(go_up.0(g.1-1(x0, x1))) -> TOP.0(check_g.0(redex_g.1-1(x0, x1))) TOP.0(go_up.1(f.0-0(x0, x0))) -> TOP.0(go_up.0(g.0-0(x0, x0))) TOP.0(go_up.0(f.1-1(x0, x0))) -> TOP.0(go_up.0(g.1-1(x0, x0))) The TRS R consists of the following rules: reduce.0(g.0-0(x_1, x_2)) -> check_g.0(redex_g.0-0(x_1, x_2)) reduce.0(g.0-1(x_1, x_2)) -> check_g.0(redex_g.0-1(x_1, x_2)) reduce.0(g.1-0(x_1, x_2)) -> check_g.0(redex_g.1-0(x_1, x_2)) reduce.0(g.1-1(x_1, x_2)) -> check_g.0(redex_g.1-1(x_1, x_2)) reduce.1(f.0-0(x_1, x_2)) -> check_f.0(redex_f.0-0(x_1, x_2)) reduce.1(f.0-1(x_1, x_2)) -> check_f.0(redex_f.0-1(x_1, x_2)) reduce.0(f.1-0(x_1, x_2)) -> check_f.0(redex_f.1-0(x_1, x_2)) reduce.0(f.1-1(x_1, x_2)) -> check_f.0(redex_f.1-1(x_1, x_2)) redex_f.0-0(x, x) -> result_f.0(g.0-0(x, x)) redex_f.1-1(x, x) -> result_f.0(g.1-1(x, x)) redex_f.0-0(x, y) -> result_f.0(y) redex_f.0-1(x, y) -> result_f.1(y) redex_f.1-0(x, y) -> result_f.0(y) redex_f.1-1(x, y) -> result_f.1(y) check_f.0(result_f.0(x)) -> go_up.0(x) check_f.0(result_f.1(x)) -> go_up.1(x) redex_g.0-0(x, x) -> result_g.0(f.1-0(f.0-0(x, x), x)) redex_g.1-1(x, x) -> result_g.1(f.0-1(f.1-1(x, x), x)) check_g.0(result_g.0(x)) -> go_up.0(x) check_g.0(result_g.1(x)) -> go_up.1(x) check_g.0(redex_g.0-0(x_1, x_2)) -> in_g_1.0-0(reduce.0(x_1), x_2) check_g.0(redex_g.0-1(x_1, x_2)) -> in_g_1.0-1(reduce.0(x_1), x_2) check_g.0(redex_g.1-0(x_1, x_2)) -> in_g_1.0-0(reduce.1(x_1), x_2) check_g.0(redex_g.1-1(x_1, x_2)) -> in_g_1.0-1(reduce.1(x_1), x_2) check_g.0(redex_g.0-0(x_1, x_2)) -> in_g_2.0-0(x_1, reduce.0(x_2)) check_g.0(redex_g.0-1(x_1, x_2)) -> in_g_2.0-0(x_1, reduce.1(x_2)) check_g.0(redex_g.1-0(x_1, x_2)) -> in_g_2.1-0(x_1, reduce.0(x_2)) check_g.0(redex_g.1-1(x_1, x_2)) -> in_g_2.1-0(x_1, reduce.1(x_2)) in_g_2.0-0(x_1, go_up.0(x_2)) -> go_up.0(g.0-0(x_1, x_2)) in_g_2.0-0(x_1, go_up.1(x_2)) -> go_up.0(g.0-1(x_1, x_2)) in_g_2.1-0(x_1, go_up.0(x_2)) -> go_up.0(g.1-0(x_1, x_2)) in_g_2.1-0(x_1, go_up.1(x_2)) -> go_up.0(g.1-1(x_1, x_2)) in_g_1.0-0(go_up.0(x_1), x_2) -> go_up.0(g.0-0(x_1, x_2)) in_g_1.0-1(go_up.0(x_1), x_2) -> go_up.0(g.0-1(x_1, x_2)) in_g_1.0-0(go_up.1(x_1), x_2) -> go_up.0(g.1-0(x_1, x_2)) in_g_1.0-1(go_up.1(x_1), x_2) -> go_up.0(g.1-1(x_1, x_2)) The set Q consists of the following terms: reduce.0(g.0-0(x0, x1)) reduce.0(g.0-1(x0, x1)) reduce.0(g.1-0(x0, x1)) reduce.0(g.1-1(x0, x1)) reduce.1(f.0-0(x0, x1)) reduce.1(f.0-1(x0, x1)) reduce.0(f.1-0(x0, x1)) reduce.0(f.1-1(x0, x1)) redex_g.0-0(x0, x0) redex_g.1-1(x0, x0) redex_f.0-0(x0, x1) redex_f.0-1(x0, x1) redex_f.1-0(x0, x1) redex_f.1-1(x0, x1) check_g.0(result_g.0(x0)) check_g.0(result_g.1(x0)) check_f.0(result_f.0(x0)) check_f.0(result_f.1(x0)) check_g.0(redex_g.0-0(x0, x1)) check_g.0(redex_g.0-1(x0, x1)) check_g.0(redex_g.1-0(x0, x1)) check_g.0(redex_g.1-1(x0, x1)) in_g_1.0-0(go_up.0(x0), x1) in_g_1.0-1(go_up.0(x0), x1) in_g_1.0-0(go_up.1(x0), x1) in_g_1.0-1(go_up.1(x0), x1) in_g_2.0-0(x0, go_up.0(x1)) in_g_2.0-0(x0, go_up.1(x1)) in_g_2.1-0(x0, go_up.0(x1)) in_g_2.1-0(x0, go_up.1(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (38) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. TOP.0(go_up.1(f.0-0(x0, x0))) -> TOP.0(go_up.0(g.0-0(x0, x0))) TOP.0(go_up.0(f.1-1(x0, x0))) -> TOP.0(go_up.0(g.1-1(x0, x0))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(TOP.0(x_1)) = x_1 POL(check_f.0(x_1)) = 0 POL(check_g.0(x_1)) = x_1 POL(f.0-0(x_1, x_2)) = 1 + x_2 POL(f.0-1(x_1, x_2)) = 0 POL(f.1-0(x_1, x_2)) = 0 POL(f.1-1(x_1, x_2)) = 1 + x_2 POL(g.0-0(x_1, x_2)) = 0 POL(g.0-1(x_1, x_2)) = 0 POL(g.1-0(x_1, x_2)) = 0 POL(g.1-1(x_1, x_2)) = 0 POL(go_up.0(x_1)) = x_1 POL(go_up.1(x_1)) = x_1 POL(in_g_1.0-0(x_1, x_2)) = 0 POL(in_g_1.0-1(x_1, x_2)) = 0 POL(in_g_2.0-0(x_1, x_2)) = 0 POL(in_g_2.1-0(x_1, x_2)) = 0 POL(redex_f.0-0(x_1, x_2)) = x_1 + x_2 POL(redex_f.0-1(x_1, x_2)) = 1 + x_1 + x_2 POL(redex_f.1-0(x_1, x_2)) = 1 + x_1 + x_2 POL(redex_f.1-1(x_1, x_2)) = x_1 POL(redex_g.0-0(x_1, x_2)) = 0 POL(redex_g.0-1(x_1, x_2)) = 0 POL(redex_g.1-0(x_1, x_2)) = 0 POL(redex_g.1-1(x_1, x_2)) = 0 POL(reduce.0(x_1)) = 0 POL(reduce.1(x_1)) = 0 POL(result_f.0(x_1)) = 1 POL(result_f.1(x_1)) = 1 + x_1 POL(result_g.0(x_1)) = x_1 POL(result_g.1(x_1)) = x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: redex_g.0-0(x, x) -> result_g.0(f.1-0(f.0-0(x, x), x)) check_g.0(result_g.0(x)) -> go_up.0(x) check_g.0(result_g.1(x)) -> go_up.1(x) check_g.0(redex_g.0-0(x_1, x_2)) -> in_g_1.0-0(reduce.0(x_1), x_2) check_g.0(redex_g.0-1(x_1, x_2)) -> in_g_1.0-1(reduce.0(x_1), x_2) check_g.0(redex_g.1-0(x_1, x_2)) -> in_g_1.0-0(reduce.1(x_1), x_2) check_g.0(redex_g.1-1(x_1, x_2)) -> in_g_1.0-1(reduce.1(x_1), x_2) check_g.0(redex_g.0-0(x_1, x_2)) -> in_g_2.0-0(x_1, reduce.0(x_2)) check_g.0(redex_g.0-1(x_1, x_2)) -> in_g_2.0-0(x_1, reduce.1(x_2)) check_g.0(redex_g.1-0(x_1, x_2)) -> in_g_2.1-0(x_1, reduce.0(x_2)) check_g.0(redex_g.1-1(x_1, x_2)) -> in_g_2.1-0(x_1, reduce.1(x_2)) redex_g.1-1(x, x) -> result_g.1(f.0-1(f.1-1(x, x), x)) in_g_1.0-0(go_up.0(x_1), x_2) -> go_up.0(g.0-0(x_1, x_2)) in_g_1.0-0(go_up.1(x_1), x_2) -> go_up.0(g.1-0(x_1, x_2)) reduce.0(g.0-0(x_1, x_2)) -> check_g.0(redex_g.0-0(x_1, x_2)) in_g_1.0-1(go_up.0(x_1), x_2) -> go_up.0(g.0-1(x_1, x_2)) in_g_1.0-1(go_up.1(x_1), x_2) -> go_up.0(g.1-1(x_1, x_2)) reduce.0(g.0-1(x_1, x_2)) -> check_g.0(redex_g.0-1(x_1, x_2)) reduce.0(g.1-0(x_1, x_2)) -> check_g.0(redex_g.1-0(x_1, x_2)) in_g_2.1-0(x_1, go_up.0(x_2)) -> go_up.0(g.1-0(x_1, x_2)) in_g_2.1-0(x_1, go_up.1(x_2)) -> go_up.0(g.1-1(x_1, x_2)) reduce.0(g.1-1(x_1, x_2)) -> check_g.0(redex_g.1-1(x_1, x_2)) in_g_2.0-0(x_1, go_up.0(x_2)) -> go_up.0(g.0-0(x_1, x_2)) in_g_2.0-0(x_1, go_up.1(x_2)) -> go_up.0(g.0-1(x_1, x_2)) ---------------------------------------- (39) Obligation: Q DP problem: The TRS P consists of the following rules: TOP.0(go_up.0(g.0-0(x0, x1))) -> TOP.0(check_g.0(redex_g.0-0(x0, x1))) TOP.0(go_up.0(g.0-1(x0, x1))) -> TOP.0(check_g.0(redex_g.0-1(x0, x1))) TOP.0(go_up.0(g.1-0(x0, x1))) -> TOP.0(check_g.0(redex_g.1-0(x0, x1))) TOP.0(go_up.0(g.1-1(x0, x1))) -> TOP.0(check_g.0(redex_g.1-1(x0, x1))) The TRS R consists of the following rules: reduce.0(g.0-0(x_1, x_2)) -> check_g.0(redex_g.0-0(x_1, x_2)) reduce.0(g.0-1(x_1, x_2)) -> check_g.0(redex_g.0-1(x_1, x_2)) reduce.0(g.1-0(x_1, x_2)) -> check_g.0(redex_g.1-0(x_1, x_2)) reduce.0(g.1-1(x_1, x_2)) -> check_g.0(redex_g.1-1(x_1, x_2)) reduce.1(f.0-0(x_1, x_2)) -> check_f.0(redex_f.0-0(x_1, x_2)) reduce.1(f.0-1(x_1, x_2)) -> check_f.0(redex_f.0-1(x_1, x_2)) reduce.0(f.1-0(x_1, x_2)) -> check_f.0(redex_f.1-0(x_1, x_2)) reduce.0(f.1-1(x_1, x_2)) -> check_f.0(redex_f.1-1(x_1, x_2)) redex_f.0-0(x, x) -> result_f.0(g.0-0(x, x)) redex_f.1-1(x, x) -> result_f.0(g.1-1(x, x)) redex_f.0-0(x, y) -> result_f.0(y) redex_f.0-1(x, y) -> result_f.1(y) redex_f.1-0(x, y) -> result_f.0(y) redex_f.1-1(x, y) -> result_f.1(y) check_f.0(result_f.0(x)) -> go_up.0(x) check_f.0(result_f.1(x)) -> go_up.1(x) redex_g.0-0(x, x) -> result_g.0(f.1-0(f.0-0(x, x), x)) redex_g.1-1(x, x) -> result_g.1(f.0-1(f.1-1(x, x), x)) check_g.0(result_g.0(x)) -> go_up.0(x) check_g.0(result_g.1(x)) -> go_up.1(x) check_g.0(redex_g.0-0(x_1, x_2)) -> in_g_1.0-0(reduce.0(x_1), x_2) check_g.0(redex_g.0-1(x_1, x_2)) -> in_g_1.0-1(reduce.0(x_1), x_2) check_g.0(redex_g.1-0(x_1, x_2)) -> in_g_1.0-0(reduce.1(x_1), x_2) check_g.0(redex_g.1-1(x_1, x_2)) -> in_g_1.0-1(reduce.1(x_1), x_2) check_g.0(redex_g.0-0(x_1, x_2)) -> in_g_2.0-0(x_1, reduce.0(x_2)) check_g.0(redex_g.0-1(x_1, x_2)) -> in_g_2.0-0(x_1, reduce.1(x_2)) check_g.0(redex_g.1-0(x_1, x_2)) -> in_g_2.1-0(x_1, reduce.0(x_2)) check_g.0(redex_g.1-1(x_1, x_2)) -> in_g_2.1-0(x_1, reduce.1(x_2)) in_g_2.0-0(x_1, go_up.0(x_2)) -> go_up.0(g.0-0(x_1, x_2)) in_g_2.0-0(x_1, go_up.1(x_2)) -> go_up.0(g.0-1(x_1, x_2)) in_g_2.1-0(x_1, go_up.0(x_2)) -> go_up.0(g.1-0(x_1, x_2)) in_g_2.1-0(x_1, go_up.1(x_2)) -> go_up.0(g.1-1(x_1, x_2)) in_g_1.0-0(go_up.0(x_1), x_2) -> go_up.0(g.0-0(x_1, x_2)) in_g_1.0-1(go_up.0(x_1), x_2) -> go_up.0(g.0-1(x_1, x_2)) in_g_1.0-0(go_up.1(x_1), x_2) -> go_up.0(g.1-0(x_1, x_2)) in_g_1.0-1(go_up.1(x_1), x_2) -> go_up.0(g.1-1(x_1, x_2)) The set Q consists of the following terms: reduce.0(g.0-0(x0, x1)) reduce.0(g.0-1(x0, x1)) reduce.0(g.1-0(x0, x1)) reduce.0(g.1-1(x0, x1)) reduce.1(f.0-0(x0, x1)) reduce.1(f.0-1(x0, x1)) reduce.0(f.1-0(x0, x1)) reduce.0(f.1-1(x0, x1)) redex_g.0-0(x0, x0) redex_g.1-1(x0, x0) redex_f.0-0(x0, x1) redex_f.0-1(x0, x1) redex_f.1-0(x0, x1) redex_f.1-1(x0, x1) check_g.0(result_g.0(x0)) check_g.0(result_g.1(x0)) check_f.0(result_f.0(x0)) check_f.0(result_f.1(x0)) check_g.0(redex_g.0-0(x0, x1)) check_g.0(redex_g.0-1(x0, x1)) check_g.0(redex_g.1-0(x0, x1)) check_g.0(redex_g.1-1(x0, x1)) in_g_1.0-0(go_up.0(x0), x1) in_g_1.0-1(go_up.0(x0), x1) in_g_1.0-0(go_up.1(x0), x1) in_g_1.0-1(go_up.1(x0), x1) in_g_2.0-0(x0, go_up.0(x1)) in_g_2.0-0(x0, go_up.1(x1)) in_g_2.1-0(x0, go_up.0(x1)) in_g_2.1-0(x0, go_up.1(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (40) PisEmptyProof (SOUND) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (41) TRUE ---------------------------------------- (42) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(g(x0, x1))) -> TOP(check_g(redex_g(x0, x1))) The TRS R consists of the following rules: reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_f(result_f(x)) -> go_up(x) redex_g(x, x) -> result_g(f(f(x, x), x)) check_g(result_g(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) The set Q consists of the following terms: reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (43) 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. ---------------------------------------- (44) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) The TRS R consists of the following rules: reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_f(result_f(x)) -> go_up(x) redex_g(x, x) -> result_g(f(f(x, x), x)) check_g(result_g(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) The set Q consists of the following terms: top(go_up(x0)) reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (45) 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(go_up(x0)) in_f_1(go_up(x0), x1) in_f_2(x0, go_up(x1)) ---------------------------------------- (46) Obligation: Q DP problem: The TRS P consists of the following rules: TOP(go_up(x)) -> TOP(reduce(x)) The TRS R consists of the following rules: reduce(g(x_1, x_2)) -> check_g(redex_g(x_1, x_2)) reduce(f(x_1, x_2)) -> check_f(redex_f(x_1, x_2)) redex_f(x, x) -> result_f(g(x, x)) redex_f(x, y) -> result_f(y) check_f(result_f(x)) -> go_up(x) redex_g(x, x) -> result_g(f(f(x, x), x)) check_g(result_g(x)) -> go_up(x) check_g(redex_g(x_1, x_2)) -> in_g_1(reduce(x_1), x_2) check_g(redex_g(x_1, x_2)) -> in_g_2(x_1, reduce(x_2)) in_g_2(x_1, go_up(x_2)) -> go_up(g(x_1, x_2)) in_g_1(go_up(x_1), x_2) -> go_up(g(x_1, x_2)) The set Q consists of the following terms: reduce(g(x0, x1)) reduce(f(x0, x1)) redex_g(x0, x0) redex_f(x0, x1) check_g(result_g(x0)) check_f(result_f(x0)) check_g(redex_g(x0, x1)) in_g_1(go_up(x0), x1) in_g_2(x0, go_up(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (47) Trivial-Transformation (SOUND) We applied the Trivial transformation to transform the outermost TRS to a standard TRS. ---------------------------------------- (48) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: g(x, x) -> f(f(x, x), x) f(x, x) -> g(x, x) f(x, y) -> y Q is empty. ---------------------------------------- (49) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (50) Obligation: Q DP problem: The TRS P consists of the following rules: G(x, x) -> F(f(x, x), x) G(x, x) -> F(x, x) F(x, x) -> G(x, x) The TRS R consists of the following rules: g(x, x) -> f(f(x, x), x) f(x, x) -> g(x, x) f(x, y) -> y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (51) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (52) Obligation: Q DP problem: The TRS P consists of the following rules: G(x, x) -> F(f(x, x), x) G(x, x) -> F(x, x) F(x, x) -> G(x, x) The TRS R consists of the following rules: g(x, x) -> f(f(x, x), x) f(x, x) -> g(x, x) f(x, y) -> y The set Q consists of the following terms: g(x0, x0) f(x0, x1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (53) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule G(x, x) -> F(f(x, x), x) at position [0] we obtained the following new rules [LPAR04]: (G(x0, x0) -> F(g(x0, x0), x0),G(x0, x0) -> F(g(x0, x0), x0)) (G(x0, x0) -> F(x0, x0),G(x, x) -> F(x, x)) ---------------------------------------- (54) Obligation: Q DP problem: The TRS P consists of the following rules: G(x, x) -> F(x, x) F(x, x) -> G(x, x) G(x0, x0) -> F(g(x0, x0), x0) The TRS R consists of the following rules: g(x, x) -> f(f(x, x), x) f(x, x) -> g(x, x) f(x, y) -> y The set Q consists of the following terms: g(x0, x0) f(x0, x1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (55) MNOCProof (EQUIVALENT) We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set. ---------------------------------------- (56) Obligation: Q DP problem: The TRS P consists of the following rules: G(x, x) -> F(x, x) F(x, x) -> G(x, x) G(x0, x0) -> F(g(x0, x0), x0) The TRS R consists of the following rules: g(x, x) -> f(f(x, x), x) f(x, x) -> g(x, x) f(x, y) -> y Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (57) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the right: s = G(x', x') evaluates to t =G(x', x') Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence G(x', x') -> F(x', x') with rule G(x, x) -> F(x, x) and matcher [x / x']. F(x', x') -> G(x', x') with rule F(x'', x'') -> G(x'', x'') at position [] and matcher [x'' / x'] Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (58) NO