/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 157 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 34 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 42 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) AND (9) QDP (10) UsableRulesProof [EQUIVALENT, 0 ms] (11) QDP (12) QReductionProof [EQUIVALENT, 0 ms] (13) QDP (14) QDPSizeChangeProof [EQUIVALENT, 0 ms] (15) YES (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) QReductionProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] (22) YES (23) QDP (24) UsableRulesProof [EQUIVALENT, 0 ms] (25) QDP (26) QReductionProof [EQUIVALENT, 0 ms] (27) QDP (28) QDPSizeChangeProof [EQUIVALENT, 0 ms] (29) YES (30) QDP (31) UsableRulesProof [EQUIVALENT, 0 ms] (32) QDP (33) QReductionProof [EQUIVALENT, 0 ms] (34) QDP (35) QDPSizeChangeProof [EQUIVALENT, 0 ms] (36) YES (37) QDP (38) UsableRulesProof [EQUIVALENT, 0 ms] (39) QDP (40) QReductionProof [EQUIVALENT, 0 ms] (41) QDP (42) QDPSizeChangeProof [EQUIVALENT, 0 ms] (43) YES (44) QDP (45) MRRProof [EQUIVALENT, 44 ms] (46) QDP (47) MRRProof [EQUIVALENT, 38 ms] (48) QDP (49) QDPOrderProof [EQUIVALENT, 128 ms] (50) QDP (51) DependencyGraphProof [EQUIVALENT, 0 ms] (52) QDP (53) QDPOrderProof [EQUIVALENT, 138 ms] (54) QDP (55) DependencyGraphProof [EQUIVALENT, 0 ms] (56) AND (57) QDP (58) QDPOrderProof [EQUIVALENT, 0 ms] (59) QDP (60) PisEmptyProof [EQUIVALENT, 0 ms] (61) YES (62) QDP (63) UsableRulesProof [EQUIVALENT, 0 ms] (64) QDP (65) QReductionProof [EQUIVALENT, 0 ms] (66) QDP (67) QDPSizeChangeProof [EQUIVALENT, 0 ms] (68) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) active(head(cons(X, XS))) -> mark(X) active(tail(cons(X, XS))) -> mark(XS) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(active(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + x_2 POL(head(x_1)) = 1 + x_1 POL(incr(x_1)) = 2*x_1 POL(mark(x_1)) = x_1 POL(nats) = 0 POL(odds) = 0 POL(pairs) = 0 POL(s(x_1)) = x_1 POL(tail(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: active(head(cons(X, XS))) -> mark(X) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) active(tail(cons(X, XS))) -> mark(XS) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(active(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)) = 2*x_1 POL(mark(x_1)) = x_1 POL(nats) = 0 POL(odds) = 0 POL(pairs) = 0 POL(s(x_1)) = x_1 POL(tail(x_1)) = 1 + 2*x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: active(tail(cons(X, XS))) -> mark(XS) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(nats) -> MARK(cons(0, incr(nats))) ACTIVE(nats) -> CONS(0, incr(nats)) ACTIVE(nats) -> INCR(nats) ACTIVE(pairs) -> MARK(cons(0, incr(odds))) ACTIVE(pairs) -> CONS(0, incr(odds)) ACTIVE(pairs) -> INCR(odds) ACTIVE(odds) -> MARK(incr(pairs)) ACTIVE(odds) -> INCR(pairs) ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS))) ACTIVE(incr(cons(X, XS))) -> CONS(s(X), incr(XS)) ACTIVE(incr(cons(X, XS))) -> S(X) ACTIVE(incr(cons(X, XS))) -> INCR(XS) MARK(nats) -> ACTIVE(nats) MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) MARK(cons(X1, X2)) -> CONS(mark(X1), X2) MARK(cons(X1, X2)) -> MARK(X1) MARK(0) -> ACTIVE(0) MARK(incr(X)) -> ACTIVE(incr(mark(X))) MARK(incr(X)) -> INCR(mark(X)) MARK(incr(X)) -> MARK(X) MARK(pairs) -> ACTIVE(pairs) MARK(odds) -> ACTIVE(odds) MARK(s(X)) -> ACTIVE(s(mark(X))) MARK(s(X)) -> S(mark(X)) MARK(s(X)) -> MARK(X) MARK(head(X)) -> ACTIVE(head(mark(X))) MARK(head(X)) -> HEAD(mark(X)) MARK(head(X)) -> MARK(X) MARK(tail(X)) -> ACTIVE(tail(mark(X))) MARK(tail(X)) -> TAIL(mark(X)) MARK(tail(X)) -> MARK(X) 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) INCR(mark(X)) -> INCR(X) INCR(active(X)) -> INCR(X) S(mark(X)) -> S(X) S(active(X)) -> S(X) HEAD(mark(X)) -> HEAD(X) HEAD(active(X)) -> HEAD(X) TAIL(mark(X)) -> TAIL(X) TAIL(active(X)) -> TAIL(X) The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 14 less nodes. ---------------------------------------- (8) Complex Obligation (AND) ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: TAIL(active(X)) -> TAIL(X) TAIL(mark(X)) -> TAIL(X) The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) 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. ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: TAIL(active(X)) -> TAIL(X) TAIL(mark(X)) -> TAIL(X) R is empty. The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) 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]. cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: TAIL(active(X)) -> TAIL(X) TAIL(mark(X)) -> TAIL(X) R is empty. The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) 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: *TAIL(active(X)) -> TAIL(X) The graph contains the following edges 1 > 1 *TAIL(mark(X)) -> TAIL(X) The graph contains the following edges 1 > 1 ---------------------------------------- (15) YES ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: HEAD(active(X)) -> HEAD(X) HEAD(mark(X)) -> HEAD(X) The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) 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: HEAD(active(X)) -> HEAD(X) HEAD(mark(X)) -> HEAD(X) R is empty. The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) 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]. cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: HEAD(active(X)) -> HEAD(X) HEAD(mark(X)) -> HEAD(X) R is empty. The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) 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: *HEAD(active(X)) -> HEAD(X) The graph contains the following edges 1 > 1 *HEAD(mark(X)) -> HEAD(X) The graph contains the following edges 1 > 1 ---------------------------------------- (22) YES ---------------------------------------- (23) 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(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (24) 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. ---------------------------------------- (25) 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. The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (26) 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]. cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) ---------------------------------------- (27) 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. The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) 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: *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 ---------------------------------------- (29) YES ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: INCR(active(X)) -> INCR(X) INCR(mark(X)) -> INCR(X) The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) 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. ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: INCR(active(X)) -> INCR(X) INCR(mark(X)) -> INCR(X) R is empty. The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) 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]. cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) ---------------------------------------- (34) Obligation: Q DP problem: The TRS P consists of the following rules: INCR(active(X)) -> INCR(X) INCR(mark(X)) -> INCR(X) R is empty. The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (35) 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(active(X)) -> INCR(X) The graph contains the following edges 1 > 1 *INCR(mark(X)) -> INCR(X) The graph contains the following edges 1 > 1 ---------------------------------------- (36) YES ---------------------------------------- (37) 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(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (38) 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. ---------------------------------------- (39) 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. The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (40) 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]. cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) ---------------------------------------- (41) 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. The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (42) 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 ---------------------------------------- (43) YES ---------------------------------------- (44) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS))) MARK(cons(X1, X2)) -> MARK(X1) MARK(nats) -> ACTIVE(nats) ACTIVE(nats) -> MARK(cons(0, incr(nats))) MARK(incr(X)) -> ACTIVE(incr(mark(X))) MARK(incr(X)) -> MARK(X) MARK(pairs) -> ACTIVE(pairs) ACTIVE(pairs) -> MARK(cons(0, incr(odds))) MARK(odds) -> ACTIVE(odds) ACTIVE(odds) -> MARK(incr(pairs)) MARK(s(X)) -> ACTIVE(s(mark(X))) MARK(s(X)) -> MARK(X) MARK(head(X)) -> ACTIVE(head(mark(X))) MARK(head(X)) -> MARK(X) MARK(tail(X)) -> ACTIVE(tail(mark(X))) MARK(tail(X)) -> MARK(X) The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (45) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: MARK(head(X)) -> MARK(X) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(ACTIVE(x_1)) = x_1 POL(MARK(x_1)) = x_1 POL(active(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + x_2 POL(head(x_1)) = 2 + 2*x_1 POL(incr(x_1)) = 2*x_1 POL(mark(x_1)) = x_1 POL(nats) = 0 POL(odds) = 0 POL(pairs) = 0 POL(s(x_1)) = x_1 POL(tail(x_1)) = 2*x_1 ---------------------------------------- (46) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS))) MARK(cons(X1, X2)) -> MARK(X1) MARK(nats) -> ACTIVE(nats) ACTIVE(nats) -> MARK(cons(0, incr(nats))) MARK(incr(X)) -> ACTIVE(incr(mark(X))) MARK(incr(X)) -> MARK(X) MARK(pairs) -> ACTIVE(pairs) ACTIVE(pairs) -> MARK(cons(0, incr(odds))) MARK(odds) -> ACTIVE(odds) ACTIVE(odds) -> MARK(incr(pairs)) MARK(s(X)) -> ACTIVE(s(mark(X))) MARK(s(X)) -> MARK(X) MARK(head(X)) -> ACTIVE(head(mark(X))) MARK(tail(X)) -> ACTIVE(tail(mark(X))) MARK(tail(X)) -> MARK(X) The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (47) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: MARK(tail(X)) -> MARK(X) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(ACTIVE(x_1)) = 2*x_1 POL(MARK(x_1)) = 2*x_1 POL(active(x_1)) = x_1 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(head(x_1)) = x_1 POL(incr(x_1)) = 2*x_1 POL(mark(x_1)) = x_1 POL(nats) = 0 POL(odds) = 0 POL(pairs) = 0 POL(s(x_1)) = x_1 POL(tail(x_1)) = 2 + 2*x_1 ---------------------------------------- (48) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS))) MARK(cons(X1, X2)) -> MARK(X1) MARK(nats) -> ACTIVE(nats) ACTIVE(nats) -> MARK(cons(0, incr(nats))) MARK(incr(X)) -> ACTIVE(incr(mark(X))) MARK(incr(X)) -> MARK(X) MARK(pairs) -> ACTIVE(pairs) ACTIVE(pairs) -> MARK(cons(0, incr(odds))) MARK(odds) -> ACTIVE(odds) ACTIVE(odds) -> MARK(incr(pairs)) MARK(s(X)) -> ACTIVE(s(mark(X))) MARK(s(X)) -> MARK(X) MARK(head(X)) -> ACTIVE(head(mark(X))) MARK(tail(X)) -> ACTIVE(tail(mark(X))) The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (49) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. ACTIVE(nats) -> MARK(cons(0, incr(nats))) ACTIVE(pairs) -> MARK(cons(0, incr(odds))) ACTIVE(odds) -> MARK(incr(pairs)) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. MARK(x1) = x1 cons(x1, x2) = x1 ACTIVE(x1) = x1 mark(x1) = x1 incr(x1) = x1 s(x1) = x1 nats = nats 0 = 0 pairs = pairs odds = odds head(x1) = head tail(x1) = tail active(x1) = x1 Knuth-Bendix order [KBO] with precedence:nats > 0 and weight map: nats=1 tail=2 0=1 head=2 odds=3 pairs=2 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) mark(incr(X)) -> active(incr(mark(X))) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(0) -> active(0) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(active(X)) -> incr(X) incr(mark(X)) -> incr(X) s(active(X)) -> s(X) s(mark(X)) -> s(X) head(active(X)) -> head(X) head(mark(X)) -> head(X) tail(active(X)) -> tail(X) tail(mark(X)) -> tail(X) ---------------------------------------- (50) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS))) MARK(cons(X1, X2)) -> MARK(X1) MARK(nats) -> ACTIVE(nats) MARK(incr(X)) -> ACTIVE(incr(mark(X))) MARK(incr(X)) -> MARK(X) MARK(pairs) -> ACTIVE(pairs) MARK(odds) -> ACTIVE(odds) MARK(s(X)) -> ACTIVE(s(mark(X))) MARK(s(X)) -> MARK(X) MARK(head(X)) -> ACTIVE(head(mark(X))) MARK(tail(X)) -> ACTIVE(tail(mark(X))) The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (51) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (52) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS))) MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) MARK(cons(X1, X2)) -> MARK(X1) MARK(incr(X)) -> ACTIVE(incr(mark(X))) MARK(incr(X)) -> MARK(X) MARK(s(X)) -> ACTIVE(s(mark(X))) MARK(s(X)) -> MARK(X) MARK(head(X)) -> ACTIVE(head(mark(X))) MARK(tail(X)) -> ACTIVE(tail(mark(X))) The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (53) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. MARK(cons(X1, X2)) -> MARK(X1) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. ACTIVE(x1) = x1 incr(x1) = x1 cons(x1, x2) = cons(x1) MARK(x1) = x1 s(x1) = x1 mark(x1) = x1 head(x1) = head tail(x1) = tail nats = nats active(x1) = x1 0 = 0 pairs = pairs odds = odds Knuth-Bendix order [KBO] with precedence:trivial and weight map: nats=4 tail=1 pairs=4 0=2 head=2 odds=5 cons_1=1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) mark(incr(X)) -> active(incr(mark(X))) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(0) -> active(0) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(active(X)) -> incr(X) incr(mark(X)) -> incr(X) s(active(X)) -> s(X) s(mark(X)) -> s(X) head(active(X)) -> head(X) head(mark(X)) -> head(X) tail(active(X)) -> tail(X) tail(mark(X)) -> tail(X) ---------------------------------------- (54) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS))) MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) MARK(incr(X)) -> ACTIVE(incr(mark(X))) MARK(incr(X)) -> MARK(X) MARK(s(X)) -> ACTIVE(s(mark(X))) MARK(s(X)) -> MARK(X) MARK(head(X)) -> ACTIVE(head(mark(X))) MARK(tail(X)) -> ACTIVE(tail(mark(X))) The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (55) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes. ---------------------------------------- (56) Complex Obligation (AND) ---------------------------------------- (57) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS))) The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (58) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) ACTIVE(incr(cons(X, XS))) -> MARK(cons(s(X), incr(XS))) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. MARK(x1) = MARK ACTIVE(x1) = x1 cons(x1, x2) = cons incr(x1) = incr Knuth-Bendix order [KBO] with precedence:MARK > cons and weight map: MARK=1 incr=2 cons=1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 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) ---------------------------------------- (59) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (60) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (61) YES ---------------------------------------- (62) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(s(X)) -> MARK(X) MARK(incr(X)) -> MARK(X) The TRS R consists of the following rules: active(nats) -> mark(cons(0, incr(nats))) active(pairs) -> mark(cons(0, incr(odds))) active(odds) -> mark(incr(pairs)) active(incr(cons(X, XS))) -> mark(cons(s(X), incr(XS))) mark(nats) -> active(nats) mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) mark(0) -> active(0) mark(incr(X)) -> active(incr(mark(X))) mark(pairs) -> active(pairs) mark(odds) -> active(odds) mark(s(X)) -> active(s(mark(X))) mark(head(X)) -> active(head(mark(X))) mark(tail(X)) -> active(tail(mark(X))) 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) incr(mark(X)) -> incr(X) incr(active(X)) -> incr(X) s(mark(X)) -> s(X) s(active(X)) -> s(X) head(mark(X)) -> head(X) head(active(X)) -> head(X) tail(mark(X)) -> tail(X) tail(active(X)) -> tail(X) The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (63) 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. ---------------------------------------- (64) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(s(X)) -> MARK(X) MARK(incr(X)) -> MARK(X) R is empty. The set Q consists of the following terms: active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (65) 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]. active(nats) active(pairs) active(odds) active(incr(cons(x0, x1))) active(head(cons(x0, x1))) active(tail(cons(x0, x1))) mark(nats) mark(cons(x0, x1)) mark(0) mark(incr(x0)) mark(pairs) mark(odds) mark(s(x0)) mark(head(x0)) mark(tail(x0)) cons(mark(x0), x1) cons(x0, mark(x1)) cons(active(x0), x1) cons(x0, active(x1)) head(mark(x0)) head(active(x0)) tail(mark(x0)) tail(active(x0)) ---------------------------------------- (66) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(s(X)) -> MARK(X) MARK(incr(X)) -> MARK(X) R is empty. The set Q consists of the following terms: incr(mark(x0)) incr(active(x0)) s(mark(x0)) s(active(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (67) 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(s(X)) -> MARK(X) The graph contains the following edges 1 > 1 *MARK(incr(X)) -> MARK(X) The graph contains the following edges 1 > 1 ---------------------------------------- (68) YES