/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 70 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 10 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 0 ms] (8) QTRS (9) Overlay + Local Confluence [EQUIVALENT, 0 ms] (10) QTRS (11) DependencyPairsProof [EQUIVALENT, 0 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) AND (15) QDP (16) UsableRulesProof [EQUIVALENT, 0 ms] (17) QDP (18) QReductionProof [EQUIVALENT, 0 ms] (19) QDP (20) NonTerminationLoopProof [COMPLETE, 5 ms] (21) NO (22) QDP (23) UsableRulesProof [EQUIVALENT, 0 ms] (24) QDP (25) QReductionProof [EQUIVALENT, 0 ms] (26) QDP (27) QDPSizeChangeProof [EQUIVALENT, 0 ms] (28) YES (29) QDP (30) UsableRulesProof [EQUIVALENT, 0 ms] (31) QDP (32) QReductionProof [EQUIVALENT, 0 ms] (33) QDP (34) QDPSizeChangeProof [EQUIVALENT, 0 ms] (35) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), incr(L)) adx(nil) -> nil adx(cons(X, L)) -> incr(cons(X, adx(L))) nats -> adx(zeros) zeros -> cons(0, zeros) head(cons(X, L)) -> X tail(cons(X, L)) -> L Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(adx(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + x_2 POL(head(x_1)) = 2*x_1 POL(incr(x_1)) = x_1 POL(nats) = 2 POL(nil) = 0 POL(s(x_1)) = x_1 POL(tail(x_1)) = x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: nats -> adx(zeros) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), incr(L)) adx(nil) -> nil adx(cons(X, L)) -> incr(cons(X, adx(L))) zeros -> cons(0, zeros) head(cons(X, L)) -> X tail(cons(X, L)) -> L Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(adx(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + x_2 POL(head(x_1)) = 1 + 2*x_1 POL(incr(x_1)) = x_1 POL(nil) = 0 POL(s(x_1)) = x_1 POL(tail(x_1)) = x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: head(cons(X, L)) -> X ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), incr(L)) adx(nil) -> nil adx(cons(X, L)) -> incr(cons(X, adx(L))) zeros -> cons(0, zeros) tail(cons(X, L)) -> L Q is empty. ---------------------------------------- (5) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(adx(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + x_2 POL(incr(x_1)) = x_1 POL(nil) = 0 POL(s(x_1)) = x_1 POL(tail(x_1)) = 1 + 2*x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: tail(cons(X, L)) -> L ---------------------------------------- (6) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), incr(L)) adx(nil) -> nil adx(cons(X, L)) -> incr(cons(X, adx(L))) zeros -> cons(0, zeros) Q is empty. ---------------------------------------- (7) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(adx(x_1)) = 2*x_1 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(incr(x_1)) = x_1 POL(nil) = 2 POL(s(x_1)) = x_1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: adx(nil) -> nil ---------------------------------------- (8) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), incr(L)) adx(cons(X, L)) -> incr(cons(X, adx(L))) zeros -> cons(0, zeros) Q is empty. ---------------------------------------- (9) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (10) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), incr(L)) adx(cons(X, L)) -> incr(cons(X, adx(L))) zeros -> cons(0, zeros) The set Q consists of the following terms: incr(nil) incr(cons(x0, x1)) adx(cons(x0, x1)) zeros ---------------------------------------- (11) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: INCR(cons(X, L)) -> INCR(L) ADX(cons(X, L)) -> INCR(cons(X, adx(L))) ADX(cons(X, L)) -> ADX(L) ZEROS -> ZEROS The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), incr(L)) adx(cons(X, L)) -> incr(cons(X, adx(L))) zeros -> cons(0, zeros) The set Q consists of the following terms: incr(nil) incr(cons(x0, x1)) adx(cons(x0, x1)) zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 1 less node. ---------------------------------------- (14) Complex Obligation (AND) ---------------------------------------- (15) Obligation: Q DP problem: The TRS P consists of the following rules: ZEROS -> ZEROS The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), incr(L)) adx(cons(X, L)) -> incr(cons(X, adx(L))) zeros -> cons(0, zeros) The set Q consists of the following terms: incr(nil) incr(cons(x0, x1)) adx(cons(x0, x1)) zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (16) 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. ---------------------------------------- (17) Obligation: Q DP problem: The TRS P consists of the following rules: ZEROS -> ZEROS R is empty. The set Q consists of the following terms: incr(nil) incr(cons(x0, x1)) adx(cons(x0, x1)) zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (18) 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]. incr(nil) incr(cons(x0, x1)) adx(cons(x0, x1)) zeros ---------------------------------------- (19) Obligation: Q DP problem: The TRS P consists of the following rules: ZEROS -> ZEROS R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (20) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by semiunifying a rule from P directly. s = ZEROS evaluates to t =ZEROS Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from ZEROS to ZEROS. ---------------------------------------- (21) NO ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: INCR(cons(X, L)) -> INCR(L) The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), incr(L)) adx(cons(X, L)) -> incr(cons(X, adx(L))) zeros -> cons(0, zeros) The set Q consists of the following terms: incr(nil) incr(cons(x0, x1)) adx(cons(x0, x1)) zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) 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. ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: INCR(cons(X, L)) -> INCR(L) R is empty. The set Q consists of the following terms: incr(nil) incr(cons(x0, x1)) adx(cons(x0, x1)) zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) 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]. incr(nil) incr(cons(x0, x1)) adx(cons(x0, x1)) zeros ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: INCR(cons(X, L)) -> INCR(L) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *INCR(cons(X, L)) -> INCR(L) The graph contains the following edges 1 > 1 ---------------------------------------- (28) YES ---------------------------------------- (29) Obligation: Q DP problem: The TRS P consists of the following rules: ADX(cons(X, L)) -> ADX(L) The TRS R consists of the following rules: incr(nil) -> nil incr(cons(X, L)) -> cons(s(X), incr(L)) adx(cons(X, L)) -> incr(cons(X, adx(L))) zeros -> cons(0, zeros) The set Q consists of the following terms: incr(nil) incr(cons(x0, x1)) adx(cons(x0, x1)) zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (30) 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. ---------------------------------------- (31) Obligation: Q DP problem: The TRS P consists of the following rules: ADX(cons(X, L)) -> ADX(L) R is empty. The set Q consists of the following terms: incr(nil) incr(cons(x0, x1)) adx(cons(x0, x1)) zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (32) 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]. incr(nil) incr(cons(x0, x1)) adx(cons(x0, x1)) zeros ---------------------------------------- (33) Obligation: Q DP problem: The TRS P consists of the following rules: ADX(cons(X, L)) -> ADX(L) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (34) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *ADX(cons(X, L)) -> ADX(L) The graph contains the following edges 1 > 1 ---------------------------------------- (35) YES