/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 54 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES (15) QDP (16) UsableRulesProof [EQUIVALENT, 0 ms] (17) QDP (18) QDPSizeChangeProof [EQUIVALENT, 0 ms] (19) YES (20) QDP (21) UsableRulesProof [EQUIVALENT, 0 ms] (22) QDP (23) QDPSizeChangeProof [EQUIVALENT, 0 ms] (24) YES (25) QDP (26) UsableRulesProof [EQUIVALENT, 0 ms] (27) QDP (28) QDPSizeChangeProof [EQUIVALENT, 0 ms] (29) YES (30) QDP (31) UsableRulesProof [EQUIVALENT, 0 ms] (32) QDP (33) QDPSizeChangeProof [EQUIVALENT, 0 ms] (34) YES (35) QDP (36) QDPOrderProof [EQUIVALENT, 135 ms] (37) QDP (38) QDPOrderProof [EQUIVALENT, 61 ms] (39) QDP (40) QDPOrderProof [EQUIVALENT, 34 ms] (41) QDP (42) QDPOrderProof [EQUIVALENT, 21 ms] (43) QDP (44) DependencyGraphProof [EQUIVALENT, 0 ms] (45) QDP (46) UsableRulesProof [EQUIVALENT, 0 ms] (47) QDP (48) QDPSizeChangeProof [EQUIVALENT, 0 ms] (49) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(app(nil, YS)) -> MARK(YS) ACTIVE(app(cons(X, XS), YS)) -> MARK(cons(X, app(XS, YS))) ACTIVE(app(cons(X, XS), YS)) -> CONS(X, app(XS, YS)) ACTIVE(app(cons(X, XS), YS)) -> APP(XS, YS) ACTIVE(from(X)) -> MARK(cons(X, from(s(X)))) ACTIVE(from(X)) -> CONS(X, from(s(X))) ACTIVE(from(X)) -> FROM(s(X)) ACTIVE(from(X)) -> S(X) ACTIVE(zWadr(nil, YS)) -> MARK(nil) ACTIVE(zWadr(XS, nil)) -> MARK(nil) ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) -> MARK(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) -> CONS(app(Y, cons(X, nil)), zWadr(XS, YS)) ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) -> APP(Y, cons(X, nil)) ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) -> CONS(X, nil) ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) -> ZWADR(XS, YS) ACTIVE(prefix(L)) -> MARK(cons(nil, zWadr(L, prefix(L)))) ACTIVE(prefix(L)) -> CONS(nil, zWadr(L, prefix(L))) ACTIVE(prefix(L)) -> ZWADR(L, prefix(L)) MARK(app(X1, X2)) -> ACTIVE(app(mark(X1), mark(X2))) MARK(app(X1, X2)) -> APP(mark(X1), mark(X2)) MARK(app(X1, X2)) -> MARK(X1) MARK(app(X1, X2)) -> MARK(X2) MARK(nil) -> ACTIVE(nil) MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) MARK(cons(X1, X2)) -> CONS(mark(X1), X2) MARK(cons(X1, X2)) -> MARK(X1) MARK(from(X)) -> ACTIVE(from(mark(X))) MARK(from(X)) -> FROM(mark(X)) MARK(from(X)) -> MARK(X) MARK(s(X)) -> ACTIVE(s(mark(X))) MARK(s(X)) -> S(mark(X)) MARK(s(X)) -> MARK(X) MARK(zWadr(X1, X2)) -> ACTIVE(zWadr(mark(X1), mark(X2))) MARK(zWadr(X1, X2)) -> ZWADR(mark(X1), mark(X2)) MARK(zWadr(X1, X2)) -> MARK(X1) MARK(zWadr(X1, X2)) -> MARK(X2) MARK(prefix(X)) -> ACTIVE(prefix(mark(X))) MARK(prefix(X)) -> PREFIX(mark(X)) MARK(prefix(X)) -> MARK(X) APP(mark(X1), X2) -> APP(X1, X2) APP(X1, mark(X2)) -> APP(X1, X2) APP(active(X1), X2) -> APP(X1, X2) APP(X1, active(X2)) -> APP(X1, X2) CONS(mark(X1), X2) -> CONS(X1, X2) CONS(X1, mark(X2)) -> CONS(X1, X2) CONS(active(X1), X2) -> CONS(X1, X2) CONS(X1, active(X2)) -> CONS(X1, X2) FROM(mark(X)) -> FROM(X) FROM(active(X)) -> FROM(X) S(mark(X)) -> S(X) S(active(X)) -> S(X) ZWADR(mark(X1), X2) -> ZWADR(X1, X2) ZWADR(X1, mark(X2)) -> ZWADR(X1, X2) ZWADR(active(X1), X2) -> ZWADR(X1, X2) ZWADR(X1, active(X2)) -> ZWADR(X1, X2) PREFIX(mark(X)) -> PREFIX(X) PREFIX(active(X)) -> PREFIX(X) The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 20 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: PREFIX(active(X)) -> PREFIX(X) PREFIX(mark(X)) -> PREFIX(X) The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: PREFIX(active(X)) -> PREFIX(X) PREFIX(mark(X)) -> PREFIX(X) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) 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: *PREFIX(active(X)) -> PREFIX(X) The graph contains the following edges 1 > 1 *PREFIX(mark(X)) -> PREFIX(X) The graph contains the following edges 1 > 1 ---------------------------------------- (9) YES ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: ZWADR(X1, mark(X2)) -> ZWADR(X1, X2) ZWADR(mark(X1), X2) -> ZWADR(X1, X2) ZWADR(active(X1), X2) -> ZWADR(X1, X2) ZWADR(X1, active(X2)) -> ZWADR(X1, X2) The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: ZWADR(X1, mark(X2)) -> ZWADR(X1, X2) ZWADR(mark(X1), X2) -> ZWADR(X1, X2) ZWADR(active(X1), X2) -> ZWADR(X1, X2) ZWADR(X1, active(X2)) -> ZWADR(X1, X2) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) 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: *ZWADR(X1, mark(X2)) -> ZWADR(X1, X2) The graph contains the following edges 1 >= 1, 2 > 2 *ZWADR(mark(X1), X2) -> ZWADR(X1, X2) The graph contains the following edges 1 > 1, 2 >= 2 *ZWADR(active(X1), X2) -> ZWADR(X1, X2) The graph contains the following edges 1 > 1, 2 >= 2 *ZWADR(X1, active(X2)) -> ZWADR(X1, X2) The graph contains the following edges 1 >= 1, 2 > 2 ---------------------------------------- (14) YES ---------------------------------------- (15) Obligation: Q DP problem: The TRS P consists of the following rules: S(active(X)) -> S(X) S(mark(X)) -> S(X) The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (16) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (17) Obligation: Q DP problem: The TRS P consists of the following rules: S(active(X)) -> S(X) S(mark(X)) -> S(X) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (18) 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: *S(active(X)) -> S(X) The graph contains the following edges 1 > 1 *S(mark(X)) -> S(X) The graph contains the following edges 1 > 1 ---------------------------------------- (19) YES ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: FROM(active(X)) -> FROM(X) FROM(mark(X)) -> FROM(X) The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: FROM(active(X)) -> FROM(X) FROM(mark(X)) -> FROM(X) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) 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: *FROM(active(X)) -> FROM(X) The graph contains the following edges 1 > 1 *FROM(mark(X)) -> FROM(X) The graph contains the following edges 1 > 1 ---------------------------------------- (24) YES ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: CONS(X1, mark(X2)) -> CONS(X1, X2) CONS(mark(X1), X2) -> CONS(X1, X2) CONS(active(X1), X2) -> CONS(X1, X2) CONS(X1, active(X2)) -> CONS(X1, X2) The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (26) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (27) Obligation: Q DP problem: The TRS P consists of the following rules: CONS(X1, mark(X2)) -> CONS(X1, X2) CONS(mark(X1), X2) -> CONS(X1, X2) CONS(active(X1), X2) -> CONS(X1, X2) CONS(X1, active(X2)) -> CONS(X1, X2) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (28) 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: *CONS(X1, mark(X2)) -> CONS(X1, X2) The graph contains the following edges 1 >= 1, 2 > 2 *CONS(mark(X1), X2) -> CONS(X1, X2) The graph contains the following edges 1 > 1, 2 >= 2 *CONS(active(X1), X2) -> CONS(X1, X2) The graph contains the following edges 1 > 1, 2 >= 2 *CONS(X1, active(X2)) -> CONS(X1, X2) The graph contains the following edges 1 >= 1, 2 > 2 ---------------------------------------- (29) YES ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: APP(X1, mark(X2)) -> APP(X1, X2) APP(mark(X1), X2) -> APP(X1, X2) APP(active(X1), X2) -> APP(X1, X2) APP(X1, active(X2)) -> APP(X1, X2) The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: APP(X1, mark(X2)) -> APP(X1, X2) APP(mark(X1), X2) -> APP(X1, X2) APP(active(X1), X2) -> APP(X1, X2) APP(X1, active(X2)) -> APP(X1, X2) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) 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: *APP(X1, mark(X2)) -> APP(X1, X2) The graph contains the following edges 1 >= 1, 2 > 2 *APP(mark(X1), X2) -> APP(X1, X2) The graph contains the following edges 1 > 1, 2 >= 2 *APP(active(X1), X2) -> APP(X1, X2) The graph contains the following edges 1 > 1, 2 >= 2 *APP(X1, active(X2)) -> APP(X1, X2) The graph contains the following edges 1 >= 1, 2 > 2 ---------------------------------------- (34) YES ---------------------------------------- (35) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(app(X1, X2)) -> ACTIVE(app(mark(X1), mark(X2))) ACTIVE(app(nil, YS)) -> MARK(YS) MARK(app(X1, X2)) -> MARK(X1) MARK(app(X1, X2)) -> MARK(X2) MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) ACTIVE(app(cons(X, XS), YS)) -> MARK(cons(X, app(XS, YS))) MARK(cons(X1, X2)) -> MARK(X1) MARK(from(X)) -> ACTIVE(from(mark(X))) ACTIVE(from(X)) -> MARK(cons(X, from(s(X)))) MARK(from(X)) -> MARK(X) MARK(s(X)) -> ACTIVE(s(mark(X))) ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) -> MARK(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) MARK(s(X)) -> MARK(X) MARK(zWadr(X1, X2)) -> ACTIVE(zWadr(mark(X1), mark(X2))) ACTIVE(prefix(L)) -> MARK(cons(nil, zWadr(L, prefix(L)))) MARK(zWadr(X1, X2)) -> MARK(X1) MARK(zWadr(X1, X2)) -> MARK(X2) MARK(prefix(X)) -> ACTIVE(prefix(mark(X))) MARK(prefix(X)) -> MARK(X) The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (36) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. ACTIVE(app(nil, YS)) -> MARK(YS) MARK(app(X1, X2)) -> MARK(X1) MARK(app(X1, X2)) -> MARK(X2) ACTIVE(app(cons(X, XS), YS)) -> MARK(cons(X, app(XS, YS))) ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) -> MARK(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) MARK(zWadr(X1, X2)) -> MARK(X1) MARK(zWadr(X1, X2)) -> MARK(X2) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. MARK(x1) = x1 app(x1, x2) = app(x1, x2) ACTIVE(x1) = x1 mark(x1) = x1 nil = nil cons(x1, x2) = x1 from(x1) = x1 s(x1) = x1 zWadr(x1, x2) = zWadr(x1, x2) prefix(x1) = x1 active(x1) = x1 Knuth-Bendix order [KBO] with precedence:trivial and weight map: zWadr_2=2 app_2=1 nil=1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) active(app(nil, YS)) -> mark(YS) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) mark(from(X)) -> active(from(mark(X))) active(from(X)) -> mark(cons(X, from(s(X)))) mark(s(X)) -> active(s(mark(X))) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(prefix(X)) -> active(prefix(mark(X))) mark(nil) -> active(nil) app(X1, mark(X2)) -> app(X1, X2) app(mark(X1), X2) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(active(X)) -> from(X) from(mark(X)) -> from(X) s(active(X)) -> s(X) s(mark(X)) -> s(X) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(active(X)) -> prefix(X) prefix(mark(X)) -> prefix(X) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) ---------------------------------------- (37) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(app(X1, X2)) -> ACTIVE(app(mark(X1), mark(X2))) MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) MARK(cons(X1, X2)) -> MARK(X1) MARK(from(X)) -> ACTIVE(from(mark(X))) ACTIVE(from(X)) -> MARK(cons(X, from(s(X)))) MARK(from(X)) -> MARK(X) MARK(s(X)) -> ACTIVE(s(mark(X))) MARK(s(X)) -> MARK(X) MARK(zWadr(X1, X2)) -> ACTIVE(zWadr(mark(X1), mark(X2))) ACTIVE(prefix(L)) -> MARK(cons(nil, zWadr(L, prefix(L)))) MARK(prefix(X)) -> ACTIVE(prefix(mark(X))) MARK(prefix(X)) -> MARK(X) The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. 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. ACTIVE(prefix(L)) -> MARK(cons(nil, zWadr(L, prefix(L)))) MARK(prefix(X)) -> MARK(X) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. MARK(x1) = x1 app(x1, x2) = app(x1, x2) ACTIVE(x1) = x1 mark(x1) = x1 cons(x1, x2) = x1 from(x1) = x1 s(x1) = x1 zWadr(x1, x2) = zWadr(x1, x2) prefix(x1) = prefix(x1) nil = nil active(x1) = x1 Knuth-Bendix order [KBO] with precedence:trivial and weight map: prefix_1=1 zWadr_2=2 app_2=1 nil=1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) active(app(nil, YS)) -> mark(YS) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) mark(from(X)) -> active(from(mark(X))) active(from(X)) -> mark(cons(X, from(s(X)))) mark(s(X)) -> active(s(mark(X))) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(prefix(X)) -> active(prefix(mark(X))) mark(nil) -> active(nil) app(X1, mark(X2)) -> app(X1, X2) app(mark(X1), X2) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(active(X)) -> from(X) from(mark(X)) -> from(X) s(active(X)) -> s(X) s(mark(X)) -> s(X) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(active(X)) -> prefix(X) prefix(mark(X)) -> prefix(X) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) ---------------------------------------- (39) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(app(X1, X2)) -> ACTIVE(app(mark(X1), mark(X2))) MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) MARK(cons(X1, X2)) -> MARK(X1) MARK(from(X)) -> ACTIVE(from(mark(X))) ACTIVE(from(X)) -> MARK(cons(X, from(s(X)))) MARK(from(X)) -> MARK(X) MARK(s(X)) -> ACTIVE(s(mark(X))) MARK(s(X)) -> MARK(X) MARK(zWadr(X1, X2)) -> ACTIVE(zWadr(mark(X1), mark(X2))) MARK(prefix(X)) -> ACTIVE(prefix(mark(X))) The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (40) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. MARK(s(X)) -> MARK(X) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. MARK(x1) = x1 app(x1, x2) = app(x1, x2) ACTIVE(x1) = x1 mark(x1) = x1 cons(x1, x2) = x1 from(x1) = x1 s(x1) = s(x1) zWadr(x1, x2) = zWadr(x1, x2) prefix(x1) = prefix active(x1) = x1 nil = nil Knuth-Bendix order [KBO] with precedence:prefix > nil and weight map: s_1=1 prefix=1 zWadr_2=2 app_2=1 nil=1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) active(app(nil, YS)) -> mark(YS) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) mark(from(X)) -> active(from(mark(X))) active(from(X)) -> mark(cons(X, from(s(X)))) mark(s(X)) -> active(s(mark(X))) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(prefix(X)) -> active(prefix(mark(X))) mark(nil) -> active(nil) app(X1, mark(X2)) -> app(X1, X2) app(mark(X1), X2) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(active(X)) -> from(X) from(mark(X)) -> from(X) s(active(X)) -> s(X) s(mark(X)) -> s(X) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(active(X)) -> prefix(X) prefix(mark(X)) -> prefix(X) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) ---------------------------------------- (41) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(app(X1, X2)) -> ACTIVE(app(mark(X1), mark(X2))) MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) MARK(cons(X1, X2)) -> MARK(X1) MARK(from(X)) -> ACTIVE(from(mark(X))) ACTIVE(from(X)) -> MARK(cons(X, from(s(X)))) MARK(from(X)) -> MARK(X) MARK(s(X)) -> ACTIVE(s(mark(X))) MARK(zWadr(X1, X2)) -> ACTIVE(zWadr(mark(X1), mark(X2))) MARK(prefix(X)) -> ACTIVE(prefix(mark(X))) The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (42) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. ACTIVE(from(X)) -> MARK(cons(X, from(s(X)))) MARK(from(X)) -> MARK(X) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. MARK(x1) = x1 app(x1, x2) = app(x1, x2) ACTIVE(x1) = x1 mark(x1) = x1 cons(x1, x2) = x1 from(x1) = from(x1) s(x1) = s zWadr(x1, x2) = zWadr(x1, x2) prefix(x1) = prefix active(x1) = x1 nil = nil Knuth-Bendix order [KBO] with precedence:trivial and weight map: s=1 prefix=3 zWadr_2=2 from_1=1 app_2=1 nil=2 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) active(app(nil, YS)) -> mark(YS) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) mark(from(X)) -> active(from(mark(X))) active(from(X)) -> mark(cons(X, from(s(X)))) mark(s(X)) -> active(s(mark(X))) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(prefix(X)) -> active(prefix(mark(X))) mark(nil) -> active(nil) app(X1, mark(X2)) -> app(X1, X2) app(mark(X1), X2) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(active(X)) -> from(X) from(mark(X)) -> from(X) s(active(X)) -> s(X) s(mark(X)) -> s(X) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(active(X)) -> prefix(X) prefix(mark(X)) -> prefix(X) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) ---------------------------------------- (43) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(app(X1, X2)) -> ACTIVE(app(mark(X1), mark(X2))) MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) MARK(cons(X1, X2)) -> MARK(X1) MARK(from(X)) -> ACTIVE(from(mark(X))) MARK(s(X)) -> ACTIVE(s(mark(X))) MARK(zWadr(X1, X2)) -> ACTIVE(zWadr(mark(X1), mark(X2))) MARK(prefix(X)) -> ACTIVE(prefix(mark(X))) The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (44) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 6 less nodes. ---------------------------------------- (45) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(cons(X1, X2)) -> MARK(X1) The TRS R consists of the following rules: active(app(nil, YS)) -> mark(YS) active(app(cons(X, XS), YS)) -> mark(cons(X, app(XS, YS))) active(from(X)) -> mark(cons(X, from(s(X)))) active(zWadr(nil, YS)) -> mark(nil) active(zWadr(XS, nil)) -> mark(nil) active(zWadr(cons(X, XS), cons(Y, YS))) -> mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS))) active(prefix(L)) -> mark(cons(nil, zWadr(L, prefix(L)))) mark(app(X1, X2)) -> active(app(mark(X1), mark(X2))) mark(nil) -> active(nil) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(from(X)) -> active(from(mark(X))) mark(s(X)) -> active(s(mark(X))) mark(zWadr(X1, X2)) -> active(zWadr(mark(X1), mark(X2))) mark(prefix(X)) -> active(prefix(mark(X))) app(mark(X1), X2) -> app(X1, X2) app(X1, mark(X2)) -> app(X1, X2) app(active(X1), X2) -> app(X1, X2) app(X1, active(X2)) -> app(X1, X2) cons(mark(X1), X2) -> cons(X1, X2) cons(X1, mark(X2)) -> cons(X1, X2) cons(active(X1), X2) -> cons(X1, X2) cons(X1, active(X2)) -> cons(X1, X2) from(mark(X)) -> from(X) from(active(X)) -> from(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) zWadr(mark(X1), X2) -> zWadr(X1, X2) zWadr(X1, mark(X2)) -> zWadr(X1, X2) zWadr(active(X1), X2) -> zWadr(X1, X2) zWadr(X1, active(X2)) -> zWadr(X1, X2) prefix(mark(X)) -> prefix(X) prefix(active(X)) -> prefix(X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (46) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (47) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(cons(X1, X2)) -> MARK(X1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (48) 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: *MARK(cons(X1, X2)) -> MARK(X1) The graph contains the following edges 1 > 1 ---------------------------------------- (49) YES