17.37/6.05 YES 17.37/6.07 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 17.37/6.07 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 17.37/6.07 17.37/6.07 17.37/6.07 Termination w.r.t. Q of the given QTRS could be proven: 17.37/6.07 17.37/6.07 (0) QTRS 17.37/6.07 (1) DependencyPairsProof [EQUIVALENT, 15 ms] 17.37/6.07 (2) QDP 17.37/6.07 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 17.37/6.07 (4) AND 17.37/6.07 (5) QDP 17.37/6.07 (6) UsableRulesProof [EQUIVALENT, 0 ms] 17.37/6.07 (7) QDP 17.37/6.07 (8) QReductionProof [EQUIVALENT, 0 ms] 17.37/6.07 (9) QDP 17.37/6.07 (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] 17.37/6.07 (11) YES 17.37/6.07 (12) QDP 17.37/6.07 (13) UsableRulesProof [EQUIVALENT, 0 ms] 17.37/6.07 (14) QDP 17.37/6.07 (15) QReductionProof [EQUIVALENT, 0 ms] 17.37/6.07 (16) QDP 17.37/6.07 (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] 17.37/6.07 (18) YES 17.37/6.07 (19) QDP 17.37/6.07 (20) UsableRulesProof [EQUIVALENT, 0 ms] 17.37/6.07 (21) QDP 17.37/6.07 (22) QReductionProof [EQUIVALENT, 0 ms] 17.37/6.07 (23) QDP 17.37/6.07 (24) QDPSizeChangeProof [EQUIVALENT, 0 ms] 17.37/6.07 (25) YES 17.37/6.07 (26) QDP 17.37/6.07 (27) UsableRulesProof [EQUIVALENT, 0 ms] 17.37/6.07 (28) QDP 17.37/6.07 (29) QReductionProof [EQUIVALENT, 0 ms] 17.37/6.07 (30) QDP 17.37/6.07 (31) QDPSizeChangeProof [EQUIVALENT, 0 ms] 17.37/6.07 (32) YES 17.37/6.07 (33) QDP 17.37/6.07 (34) UsableRulesProof [EQUIVALENT, 0 ms] 17.37/6.07 (35) QDP 17.37/6.07 (36) QReductionProof [EQUIVALENT, 0 ms] 17.37/6.07 (37) QDP 17.37/6.07 (38) QDPSizeChangeProof [EQUIVALENT, 0 ms] 17.37/6.07 (39) YES 17.37/6.07 (40) QDP 17.37/6.07 (41) QDPQMonotonicMRRProof [EQUIVALENT, 98 ms] 17.37/6.07 (42) QDP 17.37/6.07 (43) QDPQMonotonicMRRProof [EQUIVALENT, 60 ms] 17.37/6.07 (44) QDP 17.37/6.07 (45) QDPOrderProof [EQUIVALENT, 135 ms] 17.37/6.07 (46) QDP 17.37/6.07 (47) QDPOrderProof [EQUIVALENT, 180 ms] 17.37/6.07 (48) QDP 17.37/6.07 (49) DependencyGraphProof [EQUIVALENT, 0 ms] 17.37/6.07 (50) QDP 17.37/6.07 (51) QDPQMonotonicMRRProof [EQUIVALENT, 25 ms] 17.37/6.07 (52) QDP 17.37/6.07 (53) QDPOrderProof [EQUIVALENT, 112 ms] 17.37/6.07 (54) QDP 17.37/6.07 (55) DependencyGraphProof [EQUIVALENT, 0 ms] 17.37/6.07 (56) AND 17.37/6.07 (57) QDP 17.37/6.07 (58) QDPOrderProof [EQUIVALENT, 66 ms] 17.37/6.07 (59) QDP 17.37/6.07 (60) DependencyGraphProof [EQUIVALENT, 0 ms] 17.37/6.07 (61) TRUE 17.37/6.07 (62) QDP 17.37/6.07 (63) UsableRulesProof [EQUIVALENT, 0 ms] 17.37/6.07 (64) QDP 17.37/6.07 (65) QReductionProof [EQUIVALENT, 0 ms] 17.37/6.07 (66) QDP 17.37/6.07 (67) QDPSizeChangeProof [EQUIVALENT, 0 ms] 17.37/6.07 (68) YES 17.37/6.07 17.37/6.07 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (0) 17.37/6.07 Obligation: 17.37/6.07 Q restricted rewrite system: 17.37/6.07 The TRS R consists of the following rules: 17.37/6.07 17.37/6.07 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.37/6.07 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.37/6.07 active(from(X)) -> mark(cons(X, from(s(X)))) 17.37/6.07 mark(2nd(X)) -> active(2nd(mark(X))) 17.37/6.07 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.37/6.07 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.37/6.07 mark(from(X)) -> active(from(mark(X))) 17.37/6.07 mark(s(X)) -> active(s(mark(X))) 17.37/6.07 2nd(mark(X)) -> 2nd(X) 17.37/6.07 2nd(active(X)) -> 2nd(X) 17.37/6.07 cons1(mark(X1), X2) -> cons1(X1, X2) 17.37/6.07 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.37/6.07 cons1(active(X1), X2) -> cons1(X1, X2) 17.37/6.07 cons1(X1, active(X2)) -> cons1(X1, X2) 17.37/6.07 cons(mark(X1), X2) -> cons(X1, X2) 17.37/6.07 cons(X1, mark(X2)) -> cons(X1, X2) 17.37/6.07 cons(active(X1), X2) -> cons(X1, X2) 17.37/6.07 cons(X1, active(X2)) -> cons(X1, X2) 17.37/6.07 from(mark(X)) -> from(X) 17.37/6.07 from(active(X)) -> from(X) 17.37/6.07 s(mark(X)) -> s(X) 17.37/6.07 s(active(X)) -> s(X) 17.37/6.07 17.37/6.07 The set Q consists of the following terms: 17.37/6.07 17.37/6.07 active(2nd(cons1(x0, cons(x1, x2)))) 17.37/6.07 active(2nd(cons(x0, x1))) 17.37/6.07 active(from(x0)) 17.37/6.07 mark(2nd(x0)) 17.37/6.07 mark(cons1(x0, x1)) 17.37/6.07 mark(cons(x0, x1)) 17.37/6.07 mark(from(x0)) 17.37/6.07 mark(s(x0)) 17.37/6.07 2nd(mark(x0)) 17.37/6.07 2nd(active(x0)) 17.37/6.07 cons1(mark(x0), x1) 17.37/6.07 cons1(x0, mark(x1)) 17.37/6.07 cons1(active(x0), x1) 17.37/6.07 cons1(x0, active(x1)) 17.37/6.07 cons(mark(x0), x1) 17.37/6.07 cons(x0, mark(x1)) 17.37/6.07 cons(active(x0), x1) 17.37/6.07 cons(x0, active(x1)) 17.37/6.07 from(mark(x0)) 17.37/6.07 from(active(x0)) 17.37/6.07 s(mark(x0)) 17.37/6.07 s(active(x0)) 17.37/6.07 17.37/6.07 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (1) DependencyPairsProof (EQUIVALENT) 17.37/6.07 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (2) 17.37/6.07 Obligation: 17.37/6.07 Q DP problem: 17.37/6.07 The TRS P consists of the following rules: 17.37/6.07 17.37/6.07 ACTIVE(2nd(cons1(X, cons(Y, Z)))) -> MARK(Y) 17.37/6.07 ACTIVE(2nd(cons(X, X1))) -> MARK(2nd(cons1(X, X1))) 17.37/6.07 ACTIVE(2nd(cons(X, X1))) -> 2ND(cons1(X, X1)) 17.37/6.07 ACTIVE(2nd(cons(X, X1))) -> CONS1(X, X1) 17.37/6.07 ACTIVE(from(X)) -> MARK(cons(X, from(s(X)))) 17.37/6.07 ACTIVE(from(X)) -> CONS(X, from(s(X))) 17.37/6.07 ACTIVE(from(X)) -> FROM(s(X)) 17.37/6.07 ACTIVE(from(X)) -> S(X) 17.37/6.07 MARK(2nd(X)) -> ACTIVE(2nd(mark(X))) 17.37/6.07 MARK(2nd(X)) -> 2ND(mark(X)) 17.37/6.07 MARK(2nd(X)) -> MARK(X) 17.37/6.07 MARK(cons1(X1, X2)) -> ACTIVE(cons1(mark(X1), mark(X2))) 17.37/6.07 MARK(cons1(X1, X2)) -> CONS1(mark(X1), mark(X2)) 17.37/6.07 MARK(cons1(X1, X2)) -> MARK(X1) 17.37/6.07 MARK(cons1(X1, X2)) -> MARK(X2) 17.37/6.07 MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) 17.37/6.07 MARK(cons(X1, X2)) -> CONS(mark(X1), X2) 17.37/6.07 MARK(cons(X1, X2)) -> MARK(X1) 17.37/6.07 MARK(from(X)) -> ACTIVE(from(mark(X))) 17.37/6.07 MARK(from(X)) -> FROM(mark(X)) 17.37/6.07 MARK(from(X)) -> MARK(X) 17.37/6.07 MARK(s(X)) -> ACTIVE(s(mark(X))) 17.37/6.07 MARK(s(X)) -> S(mark(X)) 17.37/6.07 MARK(s(X)) -> MARK(X) 17.37/6.07 2ND(mark(X)) -> 2ND(X) 17.37/6.07 2ND(active(X)) -> 2ND(X) 17.37/6.07 CONS1(mark(X1), X2) -> CONS1(X1, X2) 17.37/6.07 CONS1(X1, mark(X2)) -> CONS1(X1, X2) 17.37/6.07 CONS1(active(X1), X2) -> CONS1(X1, X2) 17.37/6.07 CONS1(X1, active(X2)) -> CONS1(X1, X2) 17.37/6.07 CONS(mark(X1), X2) -> CONS(X1, X2) 17.37/6.07 CONS(X1, mark(X2)) -> CONS(X1, X2) 17.37/6.07 CONS(active(X1), X2) -> CONS(X1, X2) 17.37/6.07 CONS(X1, active(X2)) -> CONS(X1, X2) 17.37/6.07 FROM(mark(X)) -> FROM(X) 17.37/6.07 FROM(active(X)) -> FROM(X) 17.37/6.07 S(mark(X)) -> S(X) 17.37/6.07 S(active(X)) -> S(X) 17.37/6.07 17.37/6.07 The TRS R consists of the following rules: 17.37/6.07 17.37/6.07 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.37/6.07 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.37/6.07 active(from(X)) -> mark(cons(X, from(s(X)))) 17.37/6.07 mark(2nd(X)) -> active(2nd(mark(X))) 17.37/6.07 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.37/6.07 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.37/6.07 mark(from(X)) -> active(from(mark(X))) 17.37/6.07 mark(s(X)) -> active(s(mark(X))) 17.37/6.07 2nd(mark(X)) -> 2nd(X) 17.37/6.07 2nd(active(X)) -> 2nd(X) 17.37/6.07 cons1(mark(X1), X2) -> cons1(X1, X2) 17.37/6.07 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.37/6.07 cons1(active(X1), X2) -> cons1(X1, X2) 17.37/6.07 cons1(X1, active(X2)) -> cons1(X1, X2) 17.37/6.07 cons(mark(X1), X2) -> cons(X1, X2) 17.37/6.07 cons(X1, mark(X2)) -> cons(X1, X2) 17.37/6.07 cons(active(X1), X2) -> cons(X1, X2) 17.37/6.07 cons(X1, active(X2)) -> cons(X1, X2) 17.37/6.07 from(mark(X)) -> from(X) 17.37/6.07 from(active(X)) -> from(X) 17.37/6.07 s(mark(X)) -> s(X) 17.37/6.07 s(active(X)) -> s(X) 17.37/6.07 17.37/6.07 The set Q consists of the following terms: 17.37/6.07 17.37/6.07 active(2nd(cons1(x0, cons(x1, x2)))) 17.37/6.07 active(2nd(cons(x0, x1))) 17.37/6.07 active(from(x0)) 17.37/6.07 mark(2nd(x0)) 17.37/6.07 mark(cons1(x0, x1)) 17.37/6.07 mark(cons(x0, x1)) 17.37/6.07 mark(from(x0)) 17.37/6.07 mark(s(x0)) 17.37/6.07 2nd(mark(x0)) 17.37/6.07 2nd(active(x0)) 17.37/6.07 cons1(mark(x0), x1) 17.37/6.07 cons1(x0, mark(x1)) 17.37/6.07 cons1(active(x0), x1) 17.37/6.07 cons1(x0, active(x1)) 17.37/6.07 cons(mark(x0), x1) 17.37/6.07 cons(x0, mark(x1)) 17.37/6.07 cons(active(x0), x1) 17.37/6.07 cons(x0, active(x1)) 17.37/6.07 from(mark(x0)) 17.37/6.07 from(active(x0)) 17.37/6.07 s(mark(x0)) 17.37/6.07 s(active(x0)) 17.37/6.07 17.37/6.07 We have to consider all minimal (P,Q,R)-chains. 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (3) DependencyGraphProof (EQUIVALENT) 17.37/6.07 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 10 less nodes. 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (4) 17.37/6.07 Complex Obligation (AND) 17.37/6.07 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (5) 17.37/6.07 Obligation: 17.37/6.07 Q DP problem: 17.37/6.07 The TRS P consists of the following rules: 17.37/6.07 17.37/6.07 S(active(X)) -> S(X) 17.37/6.07 S(mark(X)) -> S(X) 17.37/6.07 17.37/6.07 The TRS R consists of the following rules: 17.37/6.07 17.37/6.07 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.37/6.07 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.37/6.07 active(from(X)) -> mark(cons(X, from(s(X)))) 17.37/6.07 mark(2nd(X)) -> active(2nd(mark(X))) 17.37/6.07 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.37/6.07 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.37/6.07 mark(from(X)) -> active(from(mark(X))) 17.37/6.07 mark(s(X)) -> active(s(mark(X))) 17.37/6.07 2nd(mark(X)) -> 2nd(X) 17.37/6.07 2nd(active(X)) -> 2nd(X) 17.37/6.07 cons1(mark(X1), X2) -> cons1(X1, X2) 17.37/6.07 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.37/6.07 cons1(active(X1), X2) -> cons1(X1, X2) 17.37/6.07 cons1(X1, active(X2)) -> cons1(X1, X2) 17.37/6.07 cons(mark(X1), X2) -> cons(X1, X2) 17.37/6.07 cons(X1, mark(X2)) -> cons(X1, X2) 17.37/6.07 cons(active(X1), X2) -> cons(X1, X2) 17.37/6.07 cons(X1, active(X2)) -> cons(X1, X2) 17.37/6.07 from(mark(X)) -> from(X) 17.37/6.07 from(active(X)) -> from(X) 17.37/6.07 s(mark(X)) -> s(X) 17.37/6.07 s(active(X)) -> s(X) 17.37/6.07 17.37/6.07 The set Q consists of the following terms: 17.37/6.07 17.37/6.07 active(2nd(cons1(x0, cons(x1, x2)))) 17.37/6.07 active(2nd(cons(x0, x1))) 17.37/6.07 active(from(x0)) 17.37/6.07 mark(2nd(x0)) 17.37/6.07 mark(cons1(x0, x1)) 17.37/6.07 mark(cons(x0, x1)) 17.37/6.07 mark(from(x0)) 17.37/6.07 mark(s(x0)) 17.37/6.07 2nd(mark(x0)) 17.37/6.07 2nd(active(x0)) 17.37/6.07 cons1(mark(x0), x1) 17.37/6.07 cons1(x0, mark(x1)) 17.37/6.07 cons1(active(x0), x1) 17.37/6.07 cons1(x0, active(x1)) 17.37/6.07 cons(mark(x0), x1) 17.37/6.07 cons(x0, mark(x1)) 17.37/6.07 cons(active(x0), x1) 17.37/6.07 cons(x0, active(x1)) 17.37/6.07 from(mark(x0)) 17.37/6.07 from(active(x0)) 17.37/6.07 s(mark(x0)) 17.37/6.07 s(active(x0)) 17.37/6.07 17.37/6.07 We have to consider all minimal (P,Q,R)-chains. 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (6) UsableRulesProof (EQUIVALENT) 17.37/6.07 As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (7) 17.37/6.07 Obligation: 17.37/6.07 Q DP problem: 17.37/6.07 The TRS P consists of the following rules: 17.37/6.07 17.37/6.07 S(active(X)) -> S(X) 17.37/6.07 S(mark(X)) -> S(X) 17.37/6.07 17.37/6.07 R is empty. 17.37/6.07 The set Q consists of the following terms: 17.37/6.07 17.37/6.07 active(2nd(cons1(x0, cons(x1, x2)))) 17.37/6.07 active(2nd(cons(x0, x1))) 17.37/6.07 active(from(x0)) 17.37/6.07 mark(2nd(x0)) 17.37/6.07 mark(cons1(x0, x1)) 17.37/6.07 mark(cons(x0, x1)) 17.37/6.07 mark(from(x0)) 17.37/6.07 mark(s(x0)) 17.37/6.07 2nd(mark(x0)) 17.37/6.07 2nd(active(x0)) 17.37/6.07 cons1(mark(x0), x1) 17.37/6.07 cons1(x0, mark(x1)) 17.37/6.07 cons1(active(x0), x1) 17.37/6.07 cons1(x0, active(x1)) 17.37/6.07 cons(mark(x0), x1) 17.37/6.07 cons(x0, mark(x1)) 17.37/6.07 cons(active(x0), x1) 17.37/6.07 cons(x0, active(x1)) 17.37/6.07 from(mark(x0)) 17.37/6.07 from(active(x0)) 17.37/6.07 s(mark(x0)) 17.37/6.07 s(active(x0)) 17.37/6.07 17.37/6.07 We have to consider all minimal (P,Q,R)-chains. 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (8) QReductionProof (EQUIVALENT) 17.37/6.07 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 17.37/6.07 17.37/6.07 2nd(mark(x0)) 17.37/6.07 2nd(active(x0)) 17.37/6.07 cons1(mark(x0), x1) 17.37/6.07 cons1(x0, mark(x1)) 17.37/6.07 cons1(active(x0), x1) 17.37/6.07 cons1(x0, active(x1)) 17.37/6.07 cons(mark(x0), x1) 17.37/6.07 cons(x0, mark(x1)) 17.37/6.07 cons(active(x0), x1) 17.37/6.07 cons(x0, active(x1)) 17.37/6.07 from(mark(x0)) 17.37/6.07 from(active(x0)) 17.37/6.07 s(mark(x0)) 17.37/6.07 s(active(x0)) 17.37/6.07 17.37/6.07 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (9) 17.37/6.07 Obligation: 17.37/6.07 Q DP problem: 17.37/6.07 The TRS P consists of the following rules: 17.37/6.07 17.37/6.07 S(active(X)) -> S(X) 17.37/6.07 S(mark(X)) -> S(X) 17.37/6.07 17.37/6.07 R is empty. 17.37/6.07 The set Q consists of the following terms: 17.37/6.07 17.37/6.07 active(2nd(cons1(x0, cons(x1, x2)))) 17.37/6.07 active(2nd(cons(x0, x1))) 17.37/6.07 active(from(x0)) 17.37/6.07 mark(2nd(x0)) 17.37/6.07 mark(cons1(x0, x1)) 17.37/6.07 mark(cons(x0, x1)) 17.37/6.07 mark(from(x0)) 17.37/6.07 mark(s(x0)) 17.37/6.07 17.37/6.07 We have to consider all minimal (P,Q,R)-chains. 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (10) QDPSizeChangeProof (EQUIVALENT) 17.37/6.07 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. 17.37/6.07 17.37/6.07 From the DPs we obtained the following set of size-change graphs: 17.37/6.07 *S(active(X)) -> S(X) 17.37/6.07 The graph contains the following edges 1 > 1 17.37/6.07 17.37/6.07 17.37/6.07 *S(mark(X)) -> S(X) 17.37/6.07 The graph contains the following edges 1 > 1 17.37/6.07 17.37/6.07 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (11) 17.37/6.07 YES 17.37/6.07 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (12) 17.37/6.07 Obligation: 17.37/6.07 Q DP problem: 17.37/6.07 The TRS P consists of the following rules: 17.37/6.07 17.37/6.07 FROM(active(X)) -> FROM(X) 17.37/6.07 FROM(mark(X)) -> FROM(X) 17.37/6.07 17.37/6.07 The TRS R consists of the following rules: 17.37/6.07 17.37/6.07 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.37/6.07 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.37/6.07 active(from(X)) -> mark(cons(X, from(s(X)))) 17.37/6.07 mark(2nd(X)) -> active(2nd(mark(X))) 17.37/6.07 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.37/6.07 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.37/6.07 mark(from(X)) -> active(from(mark(X))) 17.37/6.07 mark(s(X)) -> active(s(mark(X))) 17.37/6.07 2nd(mark(X)) -> 2nd(X) 17.37/6.07 2nd(active(X)) -> 2nd(X) 17.37/6.07 cons1(mark(X1), X2) -> cons1(X1, X2) 17.37/6.07 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.37/6.07 cons1(active(X1), X2) -> cons1(X1, X2) 17.37/6.07 cons1(X1, active(X2)) -> cons1(X1, X2) 17.37/6.07 cons(mark(X1), X2) -> cons(X1, X2) 17.37/6.07 cons(X1, mark(X2)) -> cons(X1, X2) 17.37/6.07 cons(active(X1), X2) -> cons(X1, X2) 17.37/6.07 cons(X1, active(X2)) -> cons(X1, X2) 17.37/6.07 from(mark(X)) -> from(X) 17.37/6.07 from(active(X)) -> from(X) 17.37/6.07 s(mark(X)) -> s(X) 17.37/6.07 s(active(X)) -> s(X) 17.37/6.07 17.37/6.07 The set Q consists of the following terms: 17.37/6.07 17.37/6.07 active(2nd(cons1(x0, cons(x1, x2)))) 17.37/6.07 active(2nd(cons(x0, x1))) 17.37/6.07 active(from(x0)) 17.37/6.07 mark(2nd(x0)) 17.37/6.07 mark(cons1(x0, x1)) 17.37/6.07 mark(cons(x0, x1)) 17.37/6.07 mark(from(x0)) 17.37/6.07 mark(s(x0)) 17.37/6.07 2nd(mark(x0)) 17.37/6.07 2nd(active(x0)) 17.37/6.07 cons1(mark(x0), x1) 17.37/6.07 cons1(x0, mark(x1)) 17.37/6.07 cons1(active(x0), x1) 17.37/6.07 cons1(x0, active(x1)) 17.37/6.07 cons(mark(x0), x1) 17.37/6.07 cons(x0, mark(x1)) 17.37/6.07 cons(active(x0), x1) 17.37/6.07 cons(x0, active(x1)) 17.37/6.07 from(mark(x0)) 17.37/6.07 from(active(x0)) 17.37/6.07 s(mark(x0)) 17.37/6.07 s(active(x0)) 17.37/6.07 17.37/6.07 We have to consider all minimal (P,Q,R)-chains. 17.37/6.07 ---------------------------------------- 17.37/6.07 17.37/6.07 (13) UsableRulesProof (EQUIVALENT) 17.37/6.07 As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. 17.37/6.08 ---------------------------------------- 17.37/6.08 17.37/6.08 (14) 17.37/6.08 Obligation: 17.37/6.08 Q DP problem: 17.37/6.08 The TRS P consists of the following rules: 17.37/6.08 17.37/6.08 FROM(active(X)) -> FROM(X) 17.37/6.08 FROM(mark(X)) -> FROM(X) 17.37/6.08 17.37/6.08 R is empty. 17.37/6.08 The set Q consists of the following terms: 17.37/6.08 17.37/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.37/6.08 active(2nd(cons(x0, x1))) 17.37/6.08 active(from(x0)) 17.37/6.08 mark(2nd(x0)) 17.37/6.08 mark(cons1(x0, x1)) 17.37/6.08 mark(cons(x0, x1)) 17.37/6.08 mark(from(x0)) 17.37/6.08 mark(s(x0)) 17.37/6.08 2nd(mark(x0)) 17.37/6.08 2nd(active(x0)) 17.37/6.08 cons1(mark(x0), x1) 17.37/6.08 cons1(x0, mark(x1)) 17.37/6.08 cons1(active(x0), x1) 17.37/6.08 cons1(x0, active(x1)) 17.37/6.08 cons(mark(x0), x1) 17.37/6.08 cons(x0, mark(x1)) 17.37/6.08 cons(active(x0), x1) 17.37/6.08 cons(x0, active(x1)) 17.37/6.08 from(mark(x0)) 17.37/6.08 from(active(x0)) 17.37/6.08 s(mark(x0)) 17.37/6.08 s(active(x0)) 17.37/6.08 17.37/6.08 We have to consider all minimal (P,Q,R)-chains. 17.37/6.08 ---------------------------------------- 17.37/6.08 17.37/6.08 (15) QReductionProof (EQUIVALENT) 17.37/6.08 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 17.37/6.08 17.37/6.08 2nd(mark(x0)) 17.37/6.08 2nd(active(x0)) 17.37/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (16) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 FROM(active(X)) -> FROM(X) 17.47/6.08 FROM(mark(X)) -> FROM(X) 17.47/6.08 17.47/6.08 R is empty. 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (17) QDPSizeChangeProof (EQUIVALENT) 17.47/6.08 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. 17.47/6.08 17.47/6.08 From the DPs we obtained the following set of size-change graphs: 17.47/6.08 *FROM(active(X)) -> FROM(X) 17.47/6.08 The graph contains the following edges 1 > 1 17.47/6.08 17.47/6.08 17.47/6.08 *FROM(mark(X)) -> FROM(X) 17.47/6.08 The graph contains the following edges 1 > 1 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (18) 17.47/6.08 YES 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (19) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 CONS(X1, mark(X2)) -> CONS(X1, X2) 17.47/6.08 CONS(mark(X1), X2) -> CONS(X1, X2) 17.47/6.08 CONS(active(X1), X2) -> CONS(X1, X2) 17.47/6.08 CONS(X1, active(X2)) -> CONS(X1, X2) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (20) UsableRulesProof (EQUIVALENT) 17.47/6.08 As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (21) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 CONS(X1, mark(X2)) -> CONS(X1, X2) 17.47/6.08 CONS(mark(X1), X2) -> CONS(X1, X2) 17.47/6.08 CONS(active(X1), X2) -> CONS(X1, X2) 17.47/6.08 CONS(X1, active(X2)) -> CONS(X1, X2) 17.47/6.08 17.47/6.08 R is empty. 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (22) QReductionProof (EQUIVALENT) 17.47/6.08 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 17.47/6.08 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (23) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 CONS(X1, mark(X2)) -> CONS(X1, X2) 17.47/6.08 CONS(mark(X1), X2) -> CONS(X1, X2) 17.47/6.08 CONS(active(X1), X2) -> CONS(X1, X2) 17.47/6.08 CONS(X1, active(X2)) -> CONS(X1, X2) 17.47/6.08 17.47/6.08 R is empty. 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (24) QDPSizeChangeProof (EQUIVALENT) 17.47/6.08 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. 17.47/6.08 17.47/6.08 From the DPs we obtained the following set of size-change graphs: 17.47/6.08 *CONS(X1, mark(X2)) -> CONS(X1, X2) 17.47/6.08 The graph contains the following edges 1 >= 1, 2 > 2 17.47/6.08 17.47/6.08 17.47/6.08 *CONS(mark(X1), X2) -> CONS(X1, X2) 17.47/6.08 The graph contains the following edges 1 > 1, 2 >= 2 17.47/6.08 17.47/6.08 17.47/6.08 *CONS(active(X1), X2) -> CONS(X1, X2) 17.47/6.08 The graph contains the following edges 1 > 1, 2 >= 2 17.47/6.08 17.47/6.08 17.47/6.08 *CONS(X1, active(X2)) -> CONS(X1, X2) 17.47/6.08 The graph contains the following edges 1 >= 1, 2 > 2 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (25) 17.47/6.08 YES 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (26) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 CONS1(X1, mark(X2)) -> CONS1(X1, X2) 17.47/6.08 CONS1(mark(X1), X2) -> CONS1(X1, X2) 17.47/6.08 CONS1(active(X1), X2) -> CONS1(X1, X2) 17.47/6.08 CONS1(X1, active(X2)) -> CONS1(X1, X2) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (27) UsableRulesProof (EQUIVALENT) 17.47/6.08 As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (28) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 CONS1(X1, mark(X2)) -> CONS1(X1, X2) 17.47/6.08 CONS1(mark(X1), X2) -> CONS1(X1, X2) 17.47/6.08 CONS1(active(X1), X2) -> CONS1(X1, X2) 17.47/6.08 CONS1(X1, active(X2)) -> CONS1(X1, X2) 17.47/6.08 17.47/6.08 R is empty. 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (29) QReductionProof (EQUIVALENT) 17.47/6.08 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 17.47/6.08 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (30) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 CONS1(X1, mark(X2)) -> CONS1(X1, X2) 17.47/6.08 CONS1(mark(X1), X2) -> CONS1(X1, X2) 17.47/6.08 CONS1(active(X1), X2) -> CONS1(X1, X2) 17.47/6.08 CONS1(X1, active(X2)) -> CONS1(X1, X2) 17.47/6.08 17.47/6.08 R is empty. 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (31) QDPSizeChangeProof (EQUIVALENT) 17.47/6.08 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. 17.47/6.08 17.47/6.08 From the DPs we obtained the following set of size-change graphs: 17.47/6.08 *CONS1(X1, mark(X2)) -> CONS1(X1, X2) 17.47/6.08 The graph contains the following edges 1 >= 1, 2 > 2 17.47/6.08 17.47/6.08 17.47/6.08 *CONS1(mark(X1), X2) -> CONS1(X1, X2) 17.47/6.08 The graph contains the following edges 1 > 1, 2 >= 2 17.47/6.08 17.47/6.08 17.47/6.08 *CONS1(active(X1), X2) -> CONS1(X1, X2) 17.47/6.08 The graph contains the following edges 1 > 1, 2 >= 2 17.47/6.08 17.47/6.08 17.47/6.08 *CONS1(X1, active(X2)) -> CONS1(X1, X2) 17.47/6.08 The graph contains the following edges 1 >= 1, 2 > 2 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (32) 17.47/6.08 YES 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (33) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 2ND(active(X)) -> 2ND(X) 17.47/6.08 2ND(mark(X)) -> 2ND(X) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (34) UsableRulesProof (EQUIVALENT) 17.47/6.08 As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (35) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 2ND(active(X)) -> 2ND(X) 17.47/6.08 2ND(mark(X)) -> 2ND(X) 17.47/6.08 17.47/6.08 R is empty. 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (36) QReductionProof (EQUIVALENT) 17.47/6.08 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 17.47/6.08 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (37) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 2ND(active(X)) -> 2ND(X) 17.47/6.08 2ND(mark(X)) -> 2ND(X) 17.47/6.08 17.47/6.08 R is empty. 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (38) QDPSizeChangeProof (EQUIVALENT) 17.47/6.08 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. 17.47/6.08 17.47/6.08 From the DPs we obtained the following set of size-change graphs: 17.47/6.08 *2ND(active(X)) -> 2ND(X) 17.47/6.08 The graph contains the following edges 1 > 1 17.47/6.08 17.47/6.08 17.47/6.08 *2ND(mark(X)) -> 2ND(X) 17.47/6.08 The graph contains the following edges 1 > 1 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (39) 17.47/6.08 YES 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (40) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 MARK(2nd(X)) -> ACTIVE(2nd(mark(X))) 17.47/6.08 ACTIVE(2nd(cons1(X, cons(Y, Z)))) -> MARK(Y) 17.47/6.08 MARK(2nd(X)) -> MARK(X) 17.47/6.08 MARK(cons1(X1, X2)) -> ACTIVE(cons1(mark(X1), mark(X2))) 17.47/6.08 ACTIVE(2nd(cons(X, X1))) -> MARK(2nd(cons1(X, X1))) 17.47/6.08 ACTIVE(from(X)) -> MARK(cons(X, from(s(X)))) 17.47/6.08 MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) 17.47/6.08 MARK(cons(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X2) 17.47/6.08 MARK(from(X)) -> ACTIVE(from(mark(X))) 17.47/6.08 MARK(from(X)) -> MARK(X) 17.47/6.08 MARK(s(X)) -> ACTIVE(s(mark(X))) 17.47/6.08 MARK(s(X)) -> MARK(X) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (41) QDPQMonotonicMRRProof (EQUIVALENT) 17.47/6.08 By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain. 17.47/6.08 17.47/6.08 Strictly oriented dependency pairs: 17.47/6.08 17.47/6.08 MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) 17.47/6.08 MARK(s(X)) -> ACTIVE(s(mark(X))) 17.47/6.08 17.47/6.08 17.47/6.08 Used ordering: Polynomial interpretation [POLO]: 17.47/6.08 17.47/6.08 POL(2nd(x_1)) = 1 17.47/6.08 POL(ACTIVE(x_1)) = x_1 17.47/6.08 POL(MARK(x_1)) = 1 17.47/6.08 POL(active(x_1)) = 0 17.47/6.08 POL(cons(x_1, x_2)) = 0 17.47/6.08 POL(cons1(x_1, x_2)) = 1 17.47/6.08 POL(from(x_1)) = 1 17.47/6.08 POL(mark(x_1)) = 0 17.47/6.08 POL(s(x_1)) = 0 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (42) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 MARK(2nd(X)) -> ACTIVE(2nd(mark(X))) 17.47/6.08 ACTIVE(2nd(cons1(X, cons(Y, Z)))) -> MARK(Y) 17.47/6.08 MARK(2nd(X)) -> MARK(X) 17.47/6.08 MARK(cons1(X1, X2)) -> ACTIVE(cons1(mark(X1), mark(X2))) 17.47/6.08 ACTIVE(2nd(cons(X, X1))) -> MARK(2nd(cons1(X, X1))) 17.47/6.08 ACTIVE(from(X)) -> MARK(cons(X, from(s(X)))) 17.47/6.08 MARK(cons(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X2) 17.47/6.08 MARK(from(X)) -> ACTIVE(from(mark(X))) 17.47/6.08 MARK(from(X)) -> MARK(X) 17.47/6.08 MARK(s(X)) -> MARK(X) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (43) QDPQMonotonicMRRProof (EQUIVALENT) 17.47/6.08 By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain. 17.47/6.08 17.47/6.08 Strictly oriented dependency pairs: 17.47/6.08 17.47/6.08 MARK(cons1(X1, X2)) -> ACTIVE(cons1(mark(X1), mark(X2))) 17.47/6.08 17.47/6.08 17.47/6.08 Used ordering: Polynomial interpretation [POLO]: 17.47/6.08 17.47/6.08 POL(2nd(x_1)) = 1 17.47/6.08 POL(ACTIVE(x_1)) = x_1 17.47/6.08 POL(MARK(x_1)) = 1 17.47/6.08 POL(active(x_1)) = x_1 17.47/6.08 POL(cons(x_1, x_2)) = 0 17.47/6.08 POL(cons1(x_1, x_2)) = 0 17.47/6.08 POL(from(x_1)) = 1 17.47/6.08 POL(mark(x_1)) = 1 17.47/6.08 POL(s(x_1)) = 0 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (44) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 MARK(2nd(X)) -> ACTIVE(2nd(mark(X))) 17.47/6.08 ACTIVE(2nd(cons1(X, cons(Y, Z)))) -> MARK(Y) 17.47/6.08 MARK(2nd(X)) -> MARK(X) 17.47/6.08 ACTIVE(2nd(cons(X, X1))) -> MARK(2nd(cons1(X, X1))) 17.47/6.08 ACTIVE(from(X)) -> MARK(cons(X, from(s(X)))) 17.47/6.08 MARK(cons(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X2) 17.47/6.08 MARK(from(X)) -> ACTIVE(from(mark(X))) 17.47/6.08 MARK(from(X)) -> MARK(X) 17.47/6.08 MARK(s(X)) -> MARK(X) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (45) QDPOrderProof (EQUIVALENT) 17.47/6.08 We use the reduction pair processor [LPAR04,JAR06]. 17.47/6.08 17.47/6.08 17.47/6.08 The following pairs can be oriented strictly and are deleted. 17.47/6.08 17.47/6.08 MARK(from(X)) -> MARK(X) 17.47/6.08 The remaining pairs can at least be oriented weakly. 17.47/6.08 Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(MARK(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(2nd(x_1)) = [[0A]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(ACTIVE(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(mark(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(cons1(x_1, x_2)) = [[-I]] + [[0A]] * x_1 + [[0A]] * x_2 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(cons(x_1, x_2)) = [[-I]] + [[0A]] * x_1 + [[0A]] * x_2 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(from(x_1)) = [[-I]] + [[1A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(s(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(active(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 17.47/6.08 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 17.47/6.08 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (46) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 MARK(2nd(X)) -> ACTIVE(2nd(mark(X))) 17.47/6.08 ACTIVE(2nd(cons1(X, cons(Y, Z)))) -> MARK(Y) 17.47/6.08 MARK(2nd(X)) -> MARK(X) 17.47/6.08 ACTIVE(2nd(cons(X, X1))) -> MARK(2nd(cons1(X, X1))) 17.47/6.08 ACTIVE(from(X)) -> MARK(cons(X, from(s(X)))) 17.47/6.08 MARK(cons(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X2) 17.47/6.08 MARK(from(X)) -> ACTIVE(from(mark(X))) 17.47/6.08 MARK(s(X)) -> MARK(X) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (47) QDPOrderProof (EQUIVALENT) 17.47/6.08 We use the reduction pair processor [LPAR04,JAR06]. 17.47/6.08 17.47/6.08 17.47/6.08 The following pairs can be oriented strictly and are deleted. 17.47/6.08 17.47/6.08 ACTIVE(2nd(cons1(X, cons(Y, Z)))) -> MARK(Y) 17.47/6.08 MARK(cons(X1, X2)) -> MARK(X1) 17.47/6.08 The remaining pairs can at least be oriented weakly. 17.47/6.08 Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(MARK(x_1)) = [[1A]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(2nd(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(ACTIVE(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(mark(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(cons1(x_1, x_2)) = [[-I]] + [[0A]] * x_1 + [[0A]] * x_2 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(cons(x_1, x_2)) = [[2A]] + [[1A]] * x_1 + [[0A]] * x_2 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(from(x_1)) = [[2A]] + [[1A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(s(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(active(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 17.47/6.08 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 17.47/6.08 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (48) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 MARK(2nd(X)) -> ACTIVE(2nd(mark(X))) 17.47/6.08 MARK(2nd(X)) -> MARK(X) 17.47/6.08 ACTIVE(2nd(cons(X, X1))) -> MARK(2nd(cons1(X, X1))) 17.47/6.08 ACTIVE(from(X)) -> MARK(cons(X, from(s(X)))) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X2) 17.47/6.08 MARK(from(X)) -> ACTIVE(from(mark(X))) 17.47/6.08 MARK(s(X)) -> MARK(X) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (49) DependencyGraphProof (EQUIVALENT) 17.47/6.08 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (50) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 ACTIVE(2nd(cons(X, X1))) -> MARK(2nd(cons1(X, X1))) 17.47/6.08 MARK(2nd(X)) -> ACTIVE(2nd(mark(X))) 17.47/6.08 MARK(2nd(X)) -> MARK(X) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X2) 17.47/6.08 MARK(from(X)) -> ACTIVE(from(mark(X))) 17.47/6.08 MARK(s(X)) -> MARK(X) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (51) QDPQMonotonicMRRProof (EQUIVALENT) 17.47/6.08 By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain. 17.47/6.08 17.47/6.08 Strictly oriented dependency pairs: 17.47/6.08 17.47/6.08 MARK(from(X)) -> ACTIVE(from(mark(X))) 17.47/6.08 17.47/6.08 17.47/6.08 Used ordering: Polynomial interpretation [POLO]: 17.47/6.08 17.47/6.08 POL(2nd(x_1)) = 1 17.47/6.08 POL(ACTIVE(x_1)) = 2*x_1 17.47/6.08 POL(MARK(x_1)) = 2 17.47/6.08 POL(active(x_1)) = 0 17.47/6.08 POL(cons(x_1, x_2)) = 0 17.47/6.08 POL(cons1(x_1, x_2)) = 0 17.47/6.08 POL(from(x_1)) = 0 17.47/6.08 POL(mark(x_1)) = 0 17.47/6.08 POL(s(x_1)) = 0 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (52) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 ACTIVE(2nd(cons(X, X1))) -> MARK(2nd(cons1(X, X1))) 17.47/6.08 MARK(2nd(X)) -> ACTIVE(2nd(mark(X))) 17.47/6.08 MARK(2nd(X)) -> MARK(X) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X2) 17.47/6.08 MARK(s(X)) -> MARK(X) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (53) QDPOrderProof (EQUIVALENT) 17.47/6.08 We use the reduction pair processor [LPAR04,JAR06]. 17.47/6.08 17.47/6.08 17.47/6.08 The following pairs can be oriented strictly and are deleted. 17.47/6.08 17.47/6.08 MARK(2nd(X)) -> MARK(X) 17.47/6.08 The remaining pairs can at least be oriented weakly. 17.47/6.08 Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(ACTIVE(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(2nd(x_1)) = [[1A]] + [[1A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(cons(x_1, x_2)) = [[-I]] + [[0A]] * x_1 + [[0A]] * x_2 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(MARK(x_1)) = [[0A]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(cons1(x_1, x_2)) = [[-I]] + [[0A]] * x_1 + [[0A]] * x_2 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(mark(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(s(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(active(x_1)) = [[-I]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 <<< 17.47/6.08 POL(from(x_1)) = [[0A]] + [[0A]] * x_1 17.47/6.08 >>> 17.47/6.08 17.47/6.08 17.47/6.08 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 17.47/6.08 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (54) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 ACTIVE(2nd(cons(X, X1))) -> MARK(2nd(cons1(X, X1))) 17.47/6.08 MARK(2nd(X)) -> ACTIVE(2nd(mark(X))) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X2) 17.47/6.08 MARK(s(X)) -> MARK(X) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (55) DependencyGraphProof (EQUIVALENT) 17.47/6.08 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (56) 17.47/6.08 Complex Obligation (AND) 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (57) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 MARK(2nd(X)) -> ACTIVE(2nd(mark(X))) 17.47/6.08 ACTIVE(2nd(cons(X, X1))) -> MARK(2nd(cons1(X, X1))) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (58) QDPOrderProof (EQUIVALENT) 17.47/6.08 We use the reduction pair processor [LPAR04,JAR06]. 17.47/6.08 17.47/6.08 17.47/6.08 The following pairs can be oriented strictly and are deleted. 17.47/6.08 17.47/6.08 MARK(2nd(X)) -> ACTIVE(2nd(mark(X))) 17.47/6.08 The remaining pairs can at least be oriented weakly. 17.47/6.08 Used ordering: Polynomial interpretation [POLO,RATPOLO]: 17.47/6.08 17.47/6.08 POL(2nd(x_1)) = [1/4] + x_1 17.47/6.08 POL(ACTIVE(x_1)) = [1/4]x_1 17.47/6.08 POL(MARK(x_1)) = [1/2]x_1 17.47/6.08 POL(active(x_1)) = x_1 17.47/6.08 POL(cons(x_1, x_2)) = [1/4] + [4]x_1 + [1/2]x_2 17.47/6.08 POL(cons1(x_1, x_2)) = [1/2]x_1 + [1/4]x_2 17.47/6.08 POL(from(x_1)) = [1] + [4]x_1 17.47/6.08 POL(mark(x_1)) = x_1 17.47/6.08 POL(s(x_1)) = 0 17.47/6.08 The value of delta used in the strict ordering is 1/16. 17.47/6.08 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 17.47/6.08 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (59) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 ACTIVE(2nd(cons(X, X1))) -> MARK(2nd(cons1(X, X1))) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (60) DependencyGraphProof (EQUIVALENT) 17.47/6.08 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (61) 17.47/6.08 TRUE 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (62) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X2) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(s(X)) -> MARK(X) 17.47/6.08 17.47/6.08 The TRS R consists of the following rules: 17.47/6.08 17.47/6.08 active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y) 17.47/6.08 active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1))) 17.47/6.08 active(from(X)) -> mark(cons(X, from(s(X)))) 17.47/6.08 mark(2nd(X)) -> active(2nd(mark(X))) 17.47/6.08 mark(cons1(X1, X2)) -> active(cons1(mark(X1), mark(X2))) 17.47/6.08 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 17.47/6.08 mark(from(X)) -> active(from(mark(X))) 17.47/6.08 mark(s(X)) -> active(s(mark(X))) 17.47/6.08 2nd(mark(X)) -> 2nd(X) 17.47/6.08 2nd(active(X)) -> 2nd(X) 17.47/6.08 cons1(mark(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, mark(X2)) -> cons1(X1, X2) 17.47/6.08 cons1(active(X1), X2) -> cons1(X1, X2) 17.47/6.08 cons1(X1, active(X2)) -> cons1(X1, X2) 17.47/6.08 cons(mark(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, mark(X2)) -> cons(X1, X2) 17.47/6.08 cons(active(X1), X2) -> cons(X1, X2) 17.47/6.08 cons(X1, active(X2)) -> cons(X1, X2) 17.47/6.08 from(mark(X)) -> from(X) 17.47/6.08 from(active(X)) -> from(X) 17.47/6.08 s(mark(X)) -> s(X) 17.47/6.08 s(active(X)) -> s(X) 17.47/6.08 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (63) UsableRulesProof (EQUIVALENT) 17.47/6.08 As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (64) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X2) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(s(X)) -> MARK(X) 17.47/6.08 17.47/6.08 R is empty. 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (65) QReductionProof (EQUIVALENT) 17.47/6.08 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 17.47/6.08 17.47/6.08 active(2nd(cons1(x0, cons(x1, x2)))) 17.47/6.08 active(2nd(cons(x0, x1))) 17.47/6.08 active(from(x0)) 17.47/6.08 mark(2nd(x0)) 17.47/6.08 mark(cons1(x0, x1)) 17.47/6.08 mark(cons(x0, x1)) 17.47/6.08 mark(from(x0)) 17.47/6.08 mark(s(x0)) 17.47/6.08 2nd(mark(x0)) 17.47/6.08 2nd(active(x0)) 17.47/6.08 cons(mark(x0), x1) 17.47/6.08 cons(x0, mark(x1)) 17.47/6.08 cons(active(x0), x1) 17.47/6.08 cons(x0, active(x1)) 17.47/6.08 from(mark(x0)) 17.47/6.08 from(active(x0)) 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (66) 17.47/6.08 Obligation: 17.47/6.08 Q DP problem: 17.47/6.08 The TRS P consists of the following rules: 17.47/6.08 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X2) 17.47/6.08 MARK(cons1(X1, X2)) -> MARK(X1) 17.47/6.08 MARK(s(X)) -> MARK(X) 17.47/6.08 17.47/6.08 R is empty. 17.47/6.08 The set Q consists of the following terms: 17.47/6.08 17.47/6.08 cons1(mark(x0), x1) 17.47/6.08 cons1(x0, mark(x1)) 17.47/6.08 cons1(active(x0), x1) 17.47/6.08 cons1(x0, active(x1)) 17.47/6.08 s(mark(x0)) 17.47/6.08 s(active(x0)) 17.47/6.08 17.47/6.08 We have to consider all minimal (P,Q,R)-chains. 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (67) QDPSizeChangeProof (EQUIVALENT) 17.47/6.08 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. 17.47/6.08 17.47/6.08 From the DPs we obtained the following set of size-change graphs: 17.47/6.08 *MARK(cons1(X1, X2)) -> MARK(X2) 17.47/6.08 The graph contains the following edges 1 > 1 17.47/6.08 17.47/6.08 17.47/6.08 *MARK(cons1(X1, X2)) -> MARK(X1) 17.47/6.08 The graph contains the following edges 1 > 1 17.47/6.08 17.47/6.08 17.47/6.08 *MARK(s(X)) -> MARK(X) 17.47/6.08 The graph contains the following edges 1 > 1 17.47/6.08 17.47/6.08 17.47/6.08 ---------------------------------------- 17.47/6.08 17.47/6.08 (68) 17.47/6.08 YES 17.47/6.12 EOF