13.78/4.63 YES 13.83/4.64 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 13.83/4.64 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 13.83/4.64 13.83/4.64 13.83/4.64 Termination w.r.t. Q of the given QTRS could be proven: 13.83/4.64 13.83/4.64 (0) QTRS 13.83/4.64 (1) DependencyPairsProof [EQUIVALENT, 55 ms] 13.83/4.64 (2) QDP 13.83/4.64 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 13.83/4.64 (4) AND 13.83/4.64 (5) QDP 13.83/4.64 (6) UsableRulesProof [EQUIVALENT, 0 ms] 13.83/4.64 (7) QDP 13.83/4.64 (8) QReductionProof [EQUIVALENT, 0 ms] 13.83/4.64 (9) QDP 13.83/4.64 (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] 13.83/4.64 (11) YES 13.83/4.64 (12) QDP 13.83/4.64 (13) UsableRulesProof [EQUIVALENT, 0 ms] 13.83/4.64 (14) QDP 13.83/4.64 (15) QReductionProof [EQUIVALENT, 0 ms] 13.83/4.64 (16) QDP 13.83/4.64 (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] 13.83/4.64 (18) YES 13.83/4.64 (19) QDP 13.83/4.64 (20) UsableRulesProof [EQUIVALENT, 0 ms] 13.83/4.64 (21) QDP 13.83/4.64 (22) QReductionProof [EQUIVALENT, 0 ms] 13.83/4.64 (23) QDP 13.83/4.64 (24) QDPSizeChangeProof [EQUIVALENT, 0 ms] 13.83/4.64 (25) YES 13.83/4.64 (26) QDP 13.83/4.64 (27) UsableRulesProof [EQUIVALENT, 0 ms] 13.83/4.64 (28) QDP 13.83/4.64 (29) QReductionProof [EQUIVALENT, 0 ms] 13.83/4.64 (30) QDP 13.83/4.64 (31) QDPSizeChangeProof [EQUIVALENT, 0 ms] 13.83/4.64 (32) YES 13.83/4.64 (33) QDP 13.83/4.64 (34) UsableRulesProof [EQUIVALENT, 0 ms] 13.83/4.64 (35) QDP 13.83/4.64 (36) QReductionProof [EQUIVALENT, 0 ms] 13.83/4.64 (37) QDP 13.83/4.64 (38) QDPSizeChangeProof [EQUIVALENT, 0 ms] 13.83/4.64 (39) YES 13.83/4.64 (40) QDP 13.83/4.64 (41) QDPOrderProof [EQUIVALENT, 30 ms] 13.83/4.64 (42) QDP 13.83/4.64 (43) QDPOrderProof [EQUIVALENT, 23 ms] 13.83/4.64 (44) QDP 13.83/4.64 (45) DependencyGraphProof [EQUIVALENT, 0 ms] 13.83/4.64 (46) QDP 13.83/4.64 (47) QDPOrderProof [EQUIVALENT, 18 ms] 13.83/4.64 (48) QDP 13.83/4.64 (49) QDPQMonotonicMRRProof [EQUIVALENT, 76 ms] 13.83/4.64 (50) QDP 13.83/4.64 (51) QDPOrderProof [EQUIVALENT, 13 ms] 13.83/4.64 (52) QDP 13.83/4.64 (53) QDPOrderProof [EQUIVALENT, 0 ms] 13.83/4.64 (54) QDP 13.83/4.64 (55) QDPQMonotonicMRRProof [EQUIVALENT, 0 ms] 13.83/4.64 (56) QDP 13.83/4.64 (57) QDPOrderProof [EQUIVALENT, 0 ms] 13.83/4.64 (58) QDP 13.83/4.64 (59) UsableRulesProof [EQUIVALENT, 0 ms] 13.83/4.64 (60) QDP 13.83/4.64 (61) QReductionProof [EQUIVALENT, 0 ms] 13.83/4.64 (62) QDP 13.83/4.64 (63) QDPSizeChangeProof [EQUIVALENT, 0 ms] 13.83/4.64 (64) YES 13.83/4.64 13.83/4.64 13.83/4.64 ---------------------------------------- 13.83/4.64 13.83/4.64 (0) 13.83/4.64 Obligation: 13.83/4.64 Q restricted rewrite system: 13.83/4.64 The TRS R consists of the following rules: 13.83/4.64 13.83/4.64 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.64 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.64 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.64 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.64 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.64 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.64 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.64 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.64 mark(0) -> active(0) 13.83/4.64 mark(s(X)) -> active(s(mark(X))) 13.83/4.64 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.64 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.64 mark(zprimes) -> active(zprimes) 13.83/4.64 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.64 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.64 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.64 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.64 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.64 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.64 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.64 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.64 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.64 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.64 s(mark(X)) -> s(X) 13.83/4.64 s(active(X)) -> s(X) 13.83/4.64 sieve(mark(X)) -> sieve(X) 13.83/4.64 sieve(active(X)) -> sieve(X) 13.83/4.64 nats(mark(X)) -> nats(X) 13.83/4.64 nats(active(X)) -> nats(X) 13.83/4.64 13.83/4.64 The set Q consists of the following terms: 13.83/4.64 13.83/4.64 active(filter(cons(x0, x1), 0, x2)) 13.83/4.64 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.64 active(sieve(cons(0, x0))) 13.83/4.64 active(sieve(cons(s(x0), x1))) 13.83/4.64 active(nats(x0)) 13.83/4.64 active(zprimes) 13.83/4.64 mark(filter(x0, x1, x2)) 13.83/4.64 mark(cons(x0, x1)) 13.83/4.64 mark(0) 13.83/4.64 mark(s(x0)) 13.83/4.64 mark(sieve(x0)) 13.83/4.64 mark(nats(x0)) 13.83/4.64 mark(zprimes) 13.83/4.64 filter(mark(x0), x1, x2) 13.83/4.64 filter(x0, mark(x1), x2) 13.83/4.64 filter(x0, x1, mark(x2)) 13.83/4.64 filter(active(x0), x1, x2) 13.83/4.64 filter(x0, active(x1), x2) 13.83/4.64 filter(x0, x1, active(x2)) 13.83/4.64 cons(mark(x0), x1) 13.83/4.64 cons(x0, mark(x1)) 13.83/4.64 cons(active(x0), x1) 13.83/4.64 cons(x0, active(x1)) 13.83/4.64 s(mark(x0)) 13.83/4.64 s(active(x0)) 13.83/4.64 sieve(mark(x0)) 13.83/4.64 sieve(active(x0)) 13.83/4.64 nats(mark(x0)) 13.83/4.64 nats(active(x0)) 13.83/4.64 13.83/4.64 13.83/4.64 ---------------------------------------- 13.83/4.64 13.83/4.64 (1) DependencyPairsProof (EQUIVALENT) 13.83/4.64 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 13.83/4.64 ---------------------------------------- 13.83/4.64 13.83/4.64 (2) 13.83/4.64 Obligation: 13.83/4.64 Q DP problem: 13.83/4.64 The TRS P consists of the following rules: 13.83/4.64 13.83/4.64 ACTIVE(filter(cons(X, Y), 0, M)) -> MARK(cons(0, filter(Y, M, M))) 13.83/4.64 ACTIVE(filter(cons(X, Y), 0, M)) -> CONS(0, filter(Y, M, M)) 13.83/4.64 ACTIVE(filter(cons(X, Y), 0, M)) -> FILTER(Y, M, M) 13.83/4.64 ACTIVE(filter(cons(X, Y), s(N), M)) -> MARK(cons(X, filter(Y, N, M))) 13.83/4.64 ACTIVE(filter(cons(X, Y), s(N), M)) -> CONS(X, filter(Y, N, M)) 13.83/4.64 ACTIVE(filter(cons(X, Y), s(N), M)) -> FILTER(Y, N, M) 13.83/4.64 ACTIVE(sieve(cons(0, Y))) -> MARK(cons(0, sieve(Y))) 13.83/4.64 ACTIVE(sieve(cons(0, Y))) -> CONS(0, sieve(Y)) 13.83/4.64 ACTIVE(sieve(cons(0, Y))) -> SIEVE(Y) 13.83/4.64 ACTIVE(sieve(cons(s(N), Y))) -> MARK(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.64 ACTIVE(sieve(cons(s(N), Y))) -> CONS(s(N), sieve(filter(Y, N, N))) 13.83/4.64 ACTIVE(sieve(cons(s(N), Y))) -> SIEVE(filter(Y, N, N)) 13.83/4.64 ACTIVE(sieve(cons(s(N), Y))) -> FILTER(Y, N, N) 13.83/4.64 ACTIVE(nats(N)) -> MARK(cons(N, nats(s(N)))) 13.83/4.64 ACTIVE(nats(N)) -> CONS(N, nats(s(N))) 13.83/4.64 ACTIVE(nats(N)) -> NATS(s(N)) 13.83/4.64 ACTIVE(nats(N)) -> S(N) 13.83/4.64 ACTIVE(zprimes) -> MARK(sieve(nats(s(s(0))))) 13.83/4.64 ACTIVE(zprimes) -> SIEVE(nats(s(s(0)))) 13.83/4.64 ACTIVE(zprimes) -> NATS(s(s(0))) 13.83/4.64 ACTIVE(zprimes) -> S(s(0)) 13.83/4.64 ACTIVE(zprimes) -> S(0) 13.83/4.64 MARK(filter(X1, X2, X3)) -> ACTIVE(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.64 MARK(filter(X1, X2, X3)) -> FILTER(mark(X1), mark(X2), mark(X3)) 13.83/4.64 MARK(filter(X1, X2, X3)) -> MARK(X1) 13.83/4.64 MARK(filter(X1, X2, X3)) -> MARK(X2) 13.83/4.64 MARK(filter(X1, X2, X3)) -> MARK(X3) 13.83/4.64 MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) 13.83/4.64 MARK(cons(X1, X2)) -> CONS(mark(X1), X2) 13.83/4.64 MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.64 MARK(0) -> ACTIVE(0) 13.83/4.64 MARK(s(X)) -> ACTIVE(s(mark(X))) 13.83/4.64 MARK(s(X)) -> S(mark(X)) 13.83/4.64 MARK(s(X)) -> MARK(X) 13.83/4.64 MARK(sieve(X)) -> ACTIVE(sieve(mark(X))) 13.83/4.64 MARK(sieve(X)) -> SIEVE(mark(X)) 13.83/4.64 MARK(sieve(X)) -> MARK(X) 13.83/4.64 MARK(nats(X)) -> ACTIVE(nats(mark(X))) 13.83/4.64 MARK(nats(X)) -> NATS(mark(X)) 13.83/4.64 MARK(nats(X)) -> MARK(X) 13.83/4.64 MARK(zprimes) -> ACTIVE(zprimes) 13.83/4.64 FILTER(mark(X1), X2, X3) -> FILTER(X1, X2, X3) 13.83/4.64 FILTER(X1, mark(X2), X3) -> FILTER(X1, X2, X3) 13.83/4.64 FILTER(X1, X2, mark(X3)) -> FILTER(X1, X2, X3) 13.83/4.64 FILTER(active(X1), X2, X3) -> FILTER(X1, X2, X3) 13.83/4.64 FILTER(X1, active(X2), X3) -> FILTER(X1, X2, X3) 13.83/4.64 FILTER(X1, X2, active(X3)) -> FILTER(X1, X2, X3) 13.83/4.64 CONS(mark(X1), X2) -> CONS(X1, X2) 13.83/4.64 CONS(X1, mark(X2)) -> CONS(X1, X2) 13.83/4.64 CONS(active(X1), X2) -> CONS(X1, X2) 13.83/4.64 CONS(X1, active(X2)) -> CONS(X1, X2) 13.83/4.64 S(mark(X)) -> S(X) 13.83/4.64 S(active(X)) -> S(X) 13.83/4.64 SIEVE(mark(X)) -> SIEVE(X) 13.83/4.64 SIEVE(active(X)) -> SIEVE(X) 13.83/4.64 NATS(mark(X)) -> NATS(X) 13.83/4.64 NATS(active(X)) -> NATS(X) 13.83/4.64 13.83/4.64 The TRS R consists of the following rules: 13.83/4.64 13.83/4.64 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.64 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.64 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.64 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.64 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.64 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.64 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.64 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.64 mark(0) -> active(0) 13.83/4.64 mark(s(X)) -> active(s(mark(X))) 13.83/4.64 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.64 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.64 mark(zprimes) -> active(zprimes) 13.83/4.64 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.64 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.64 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.64 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.64 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.64 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.64 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.64 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.64 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.64 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.64 s(mark(X)) -> s(X) 13.83/4.64 s(active(X)) -> s(X) 13.83/4.64 sieve(mark(X)) -> sieve(X) 13.83/4.64 sieve(active(X)) -> sieve(X) 13.83/4.64 nats(mark(X)) -> nats(X) 13.83/4.64 nats(active(X)) -> nats(X) 13.83/4.64 13.83/4.64 The set Q consists of the following terms: 13.83/4.64 13.83/4.64 active(filter(cons(x0, x1), 0, x2)) 13.83/4.64 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.64 active(sieve(cons(0, x0))) 13.83/4.64 active(sieve(cons(s(x0), x1))) 13.83/4.64 active(nats(x0)) 13.83/4.64 active(zprimes) 13.83/4.64 mark(filter(x0, x1, x2)) 13.83/4.64 mark(cons(x0, x1)) 13.83/4.64 mark(0) 13.83/4.64 mark(s(x0)) 13.83/4.64 mark(sieve(x0)) 13.83/4.64 mark(nats(x0)) 13.83/4.64 mark(zprimes) 13.83/4.64 filter(mark(x0), x1, x2) 13.83/4.64 filter(x0, mark(x1), x2) 13.83/4.64 filter(x0, x1, mark(x2)) 13.83/4.64 filter(active(x0), x1, x2) 13.83/4.64 filter(x0, active(x1), x2) 13.83/4.64 filter(x0, x1, active(x2)) 13.83/4.64 cons(mark(x0), x1) 13.83/4.64 cons(x0, mark(x1)) 13.83/4.64 cons(active(x0), x1) 13.83/4.64 cons(x0, active(x1)) 13.83/4.64 s(mark(x0)) 13.83/4.64 s(active(x0)) 13.83/4.64 sieve(mark(x0)) 13.83/4.64 sieve(active(x0)) 13.83/4.64 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (3) DependencyGraphProof (EQUIVALENT) 13.83/4.65 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 22 less nodes. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (4) 13.83/4.65 Complex Obligation (AND) 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (5) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 NATS(active(X)) -> NATS(X) 13.83/4.65 NATS(mark(X)) -> NATS(X) 13.83/4.65 13.83/4.65 The TRS R consists of the following rules: 13.83/4.65 13.83/4.65 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.65 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.65 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.65 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.65 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.65 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.65 mark(0) -> active(0) 13.83/4.65 mark(s(X)) -> active(s(mark(X))) 13.83/4.65 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.65 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.65 mark(zprimes) -> active(zprimes) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (6) UsableRulesProof (EQUIVALENT) 13.83/4.65 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. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (7) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 NATS(active(X)) -> NATS(X) 13.83/4.65 NATS(mark(X)) -> NATS(X) 13.83/4.65 13.83/4.65 R is empty. 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (8) QReductionProof (EQUIVALENT) 13.83/4.65 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 13.83/4.65 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (9) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 NATS(active(X)) -> NATS(X) 13.83/4.65 NATS(mark(X)) -> NATS(X) 13.83/4.65 13.83/4.65 R is empty. 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (10) QDPSizeChangeProof (EQUIVALENT) 13.83/4.65 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. 13.83/4.65 13.83/4.65 From the DPs we obtained the following set of size-change graphs: 13.83/4.65 *NATS(active(X)) -> NATS(X) 13.83/4.65 The graph contains the following edges 1 > 1 13.83/4.65 13.83/4.65 13.83/4.65 *NATS(mark(X)) -> NATS(X) 13.83/4.65 The graph contains the following edges 1 > 1 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (11) 13.83/4.65 YES 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (12) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 SIEVE(active(X)) -> SIEVE(X) 13.83/4.65 SIEVE(mark(X)) -> SIEVE(X) 13.83/4.65 13.83/4.65 The TRS R consists of the following rules: 13.83/4.65 13.83/4.65 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.65 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.65 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.65 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.65 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.65 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.65 mark(0) -> active(0) 13.83/4.65 mark(s(X)) -> active(s(mark(X))) 13.83/4.65 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.65 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.65 mark(zprimes) -> active(zprimes) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (13) UsableRulesProof (EQUIVALENT) 13.83/4.65 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. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (14) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 SIEVE(active(X)) -> SIEVE(X) 13.83/4.65 SIEVE(mark(X)) -> SIEVE(X) 13.83/4.65 13.83/4.65 R is empty. 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (15) QReductionProof (EQUIVALENT) 13.83/4.65 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 13.83/4.65 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (16) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 SIEVE(active(X)) -> SIEVE(X) 13.83/4.65 SIEVE(mark(X)) -> SIEVE(X) 13.83/4.65 13.83/4.65 R is empty. 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (17) QDPSizeChangeProof (EQUIVALENT) 13.83/4.65 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. 13.83/4.65 13.83/4.65 From the DPs we obtained the following set of size-change graphs: 13.83/4.65 *SIEVE(active(X)) -> SIEVE(X) 13.83/4.65 The graph contains the following edges 1 > 1 13.83/4.65 13.83/4.65 13.83/4.65 *SIEVE(mark(X)) -> SIEVE(X) 13.83/4.65 The graph contains the following edges 1 > 1 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (18) 13.83/4.65 YES 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (19) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 S(active(X)) -> S(X) 13.83/4.65 S(mark(X)) -> S(X) 13.83/4.65 13.83/4.65 The TRS R consists of the following rules: 13.83/4.65 13.83/4.65 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.65 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.65 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.65 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.65 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.65 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.65 mark(0) -> active(0) 13.83/4.65 mark(s(X)) -> active(s(mark(X))) 13.83/4.65 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.65 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.65 mark(zprimes) -> active(zprimes) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (20) UsableRulesProof (EQUIVALENT) 13.83/4.65 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. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (21) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 S(active(X)) -> S(X) 13.83/4.65 S(mark(X)) -> S(X) 13.83/4.65 13.83/4.65 R is empty. 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (22) QReductionProof (EQUIVALENT) 13.83/4.65 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 13.83/4.65 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (23) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 S(active(X)) -> S(X) 13.83/4.65 S(mark(X)) -> S(X) 13.83/4.65 13.83/4.65 R is empty. 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (24) QDPSizeChangeProof (EQUIVALENT) 13.83/4.65 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. 13.83/4.65 13.83/4.65 From the DPs we obtained the following set of size-change graphs: 13.83/4.65 *S(active(X)) -> S(X) 13.83/4.65 The graph contains the following edges 1 > 1 13.83/4.65 13.83/4.65 13.83/4.65 *S(mark(X)) -> S(X) 13.83/4.65 The graph contains the following edges 1 > 1 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (25) 13.83/4.65 YES 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (26) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 CONS(X1, mark(X2)) -> CONS(X1, X2) 13.83/4.65 CONS(mark(X1), X2) -> CONS(X1, X2) 13.83/4.65 CONS(active(X1), X2) -> CONS(X1, X2) 13.83/4.65 CONS(X1, active(X2)) -> CONS(X1, X2) 13.83/4.65 13.83/4.65 The TRS R consists of the following rules: 13.83/4.65 13.83/4.65 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.65 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.65 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.65 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.65 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.65 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.65 mark(0) -> active(0) 13.83/4.65 mark(s(X)) -> active(s(mark(X))) 13.83/4.65 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.65 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.65 mark(zprimes) -> active(zprimes) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (27) UsableRulesProof (EQUIVALENT) 13.83/4.65 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. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (28) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 CONS(X1, mark(X2)) -> CONS(X1, X2) 13.83/4.65 CONS(mark(X1), X2) -> CONS(X1, X2) 13.83/4.65 CONS(active(X1), X2) -> CONS(X1, X2) 13.83/4.65 CONS(X1, active(X2)) -> CONS(X1, X2) 13.83/4.65 13.83/4.65 R is empty. 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (29) QReductionProof (EQUIVALENT) 13.83/4.65 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 13.83/4.65 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (30) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 CONS(X1, mark(X2)) -> CONS(X1, X2) 13.83/4.65 CONS(mark(X1), X2) -> CONS(X1, X2) 13.83/4.65 CONS(active(X1), X2) -> CONS(X1, X2) 13.83/4.65 CONS(X1, active(X2)) -> CONS(X1, X2) 13.83/4.65 13.83/4.65 R is empty. 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (31) QDPSizeChangeProof (EQUIVALENT) 13.83/4.65 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. 13.83/4.65 13.83/4.65 From the DPs we obtained the following set of size-change graphs: 13.83/4.65 *CONS(X1, mark(X2)) -> CONS(X1, X2) 13.83/4.65 The graph contains the following edges 1 >= 1, 2 > 2 13.83/4.65 13.83/4.65 13.83/4.65 *CONS(mark(X1), X2) -> CONS(X1, X2) 13.83/4.65 The graph contains the following edges 1 > 1, 2 >= 2 13.83/4.65 13.83/4.65 13.83/4.65 *CONS(active(X1), X2) -> CONS(X1, X2) 13.83/4.65 The graph contains the following edges 1 > 1, 2 >= 2 13.83/4.65 13.83/4.65 13.83/4.65 *CONS(X1, active(X2)) -> CONS(X1, X2) 13.83/4.65 The graph contains the following edges 1 >= 1, 2 > 2 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (32) 13.83/4.65 YES 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (33) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 FILTER(X1, mark(X2), X3) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(mark(X1), X2, X3) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(X1, X2, mark(X3)) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(active(X1), X2, X3) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(X1, active(X2), X3) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(X1, X2, active(X3)) -> FILTER(X1, X2, X3) 13.83/4.65 13.83/4.65 The TRS R consists of the following rules: 13.83/4.65 13.83/4.65 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.65 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.65 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.65 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.65 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.65 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.65 mark(0) -> active(0) 13.83/4.65 mark(s(X)) -> active(s(mark(X))) 13.83/4.65 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.65 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.65 mark(zprimes) -> active(zprimes) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (34) UsableRulesProof (EQUIVALENT) 13.83/4.65 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. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (35) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 FILTER(X1, mark(X2), X3) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(mark(X1), X2, X3) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(X1, X2, mark(X3)) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(active(X1), X2, X3) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(X1, active(X2), X3) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(X1, X2, active(X3)) -> FILTER(X1, X2, X3) 13.83/4.65 13.83/4.65 R is empty. 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (36) QReductionProof (EQUIVALENT) 13.83/4.65 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 13.83/4.65 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (37) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 FILTER(X1, mark(X2), X3) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(mark(X1), X2, X3) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(X1, X2, mark(X3)) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(active(X1), X2, X3) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(X1, active(X2), X3) -> FILTER(X1, X2, X3) 13.83/4.65 FILTER(X1, X2, active(X3)) -> FILTER(X1, X2, X3) 13.83/4.65 13.83/4.65 R is empty. 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (38) QDPSizeChangeProof (EQUIVALENT) 13.83/4.65 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. 13.83/4.65 13.83/4.65 From the DPs we obtained the following set of size-change graphs: 13.83/4.65 *FILTER(X1, mark(X2), X3) -> FILTER(X1, X2, X3) 13.83/4.65 The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3 13.83/4.65 13.83/4.65 13.83/4.65 *FILTER(mark(X1), X2, X3) -> FILTER(X1, X2, X3) 13.83/4.65 The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3 13.83/4.65 13.83/4.65 13.83/4.65 *FILTER(X1, X2, mark(X3)) -> FILTER(X1, X2, X3) 13.83/4.65 The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3 13.83/4.65 13.83/4.65 13.83/4.65 *FILTER(active(X1), X2, X3) -> FILTER(X1, X2, X3) 13.83/4.65 The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3 13.83/4.65 13.83/4.65 13.83/4.65 *FILTER(X1, active(X2), X3) -> FILTER(X1, X2, X3) 13.83/4.65 The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3 13.83/4.65 13.83/4.65 13.83/4.65 *FILTER(X1, X2, active(X3)) -> FILTER(X1, X2, X3) 13.83/4.65 The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (39) 13.83/4.65 YES 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (40) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) 13.83/4.65 ACTIVE(filter(cons(X, Y), 0, M)) -> MARK(cons(0, filter(Y, M, M))) 13.83/4.65 MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.65 MARK(filter(X1, X2, X3)) -> ACTIVE(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 ACTIVE(filter(cons(X, Y), s(N), M)) -> MARK(cons(X, filter(Y, N, M))) 13.83/4.65 ACTIVE(sieve(cons(0, Y))) -> MARK(cons(0, sieve(Y))) 13.83/4.65 ACTIVE(sieve(cons(s(N), Y))) -> MARK(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 ACTIVE(nats(N)) -> MARK(cons(N, nats(s(N)))) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X1) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X2) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X3) 13.83/4.65 MARK(s(X)) -> ACTIVE(s(mark(X))) 13.83/4.65 MARK(s(X)) -> MARK(X) 13.83/4.65 MARK(sieve(X)) -> ACTIVE(sieve(mark(X))) 13.83/4.65 MARK(sieve(X)) -> MARK(X) 13.83/4.65 MARK(nats(X)) -> ACTIVE(nats(mark(X))) 13.83/4.65 MARK(nats(X)) -> MARK(X) 13.83/4.65 MARK(zprimes) -> ACTIVE(zprimes) 13.83/4.65 ACTIVE(zprimes) -> MARK(sieve(nats(s(s(0))))) 13.83/4.65 13.83/4.65 The TRS R consists of the following rules: 13.83/4.65 13.83/4.65 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.65 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.65 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.65 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.65 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.65 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.65 mark(0) -> active(0) 13.83/4.65 mark(s(X)) -> active(s(mark(X))) 13.83/4.65 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.65 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.65 mark(zprimes) -> active(zprimes) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (41) QDPOrderProof (EQUIVALENT) 13.83/4.65 We use the reduction pair processor [LPAR04,JAR06]. 13.83/4.65 13.83/4.65 13.83/4.65 The following pairs can be oriented strictly and are deleted. 13.83/4.65 13.83/4.65 MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2)) 13.83/4.65 MARK(s(X)) -> ACTIVE(s(mark(X))) 13.83/4.65 The remaining pairs can at least be oriented weakly. 13.83/4.65 Used ordering: Polynomial interpretation [POLO]: 13.83/4.65 13.83/4.65 POL(0) = 0 13.83/4.65 POL(ACTIVE(x_1)) = x_1 13.83/4.65 POL(MARK(x_1)) = 1 13.83/4.65 POL(active(x_1)) = 0 13.83/4.65 POL(cons(x_1, x_2)) = 0 13.83/4.65 POL(filter(x_1, x_2, x_3)) = 1 13.83/4.65 POL(mark(x_1)) = 0 13.83/4.65 POL(nats(x_1)) = 1 13.83/4.65 POL(s(x_1)) = 0 13.83/4.65 POL(sieve(x_1)) = 1 13.83/4.65 POL(zprimes) = 1 13.83/4.65 13.83/4.65 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 13.83/4.65 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (42) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 ACTIVE(filter(cons(X, Y), 0, M)) -> MARK(cons(0, filter(Y, M, M))) 13.83/4.65 MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.65 MARK(filter(X1, X2, X3)) -> ACTIVE(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 ACTIVE(filter(cons(X, Y), s(N), M)) -> MARK(cons(X, filter(Y, N, M))) 13.83/4.65 ACTIVE(sieve(cons(0, Y))) -> MARK(cons(0, sieve(Y))) 13.83/4.65 ACTIVE(sieve(cons(s(N), Y))) -> MARK(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 ACTIVE(nats(N)) -> MARK(cons(N, nats(s(N)))) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X1) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X2) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X3) 13.83/4.65 MARK(s(X)) -> MARK(X) 13.83/4.65 MARK(sieve(X)) -> ACTIVE(sieve(mark(X))) 13.83/4.65 MARK(sieve(X)) -> MARK(X) 13.83/4.65 MARK(nats(X)) -> ACTIVE(nats(mark(X))) 13.83/4.65 MARK(nats(X)) -> MARK(X) 13.83/4.65 MARK(zprimes) -> ACTIVE(zprimes) 13.83/4.65 ACTIVE(zprimes) -> MARK(sieve(nats(s(s(0))))) 13.83/4.65 13.83/4.65 The TRS R consists of the following rules: 13.83/4.65 13.83/4.65 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.65 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.65 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.65 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.65 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.65 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.65 mark(0) -> active(0) 13.83/4.65 mark(s(X)) -> active(s(mark(X))) 13.83/4.65 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.65 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.65 mark(zprimes) -> active(zprimes) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (43) QDPOrderProof (EQUIVALENT) 13.83/4.65 We use the reduction pair processor [LPAR04,JAR06]. 13.83/4.65 13.83/4.65 13.83/4.65 The following pairs can be oriented strictly and are deleted. 13.83/4.65 13.83/4.65 ACTIVE(zprimes) -> MARK(sieve(nats(s(s(0))))) 13.83/4.65 The remaining pairs can at least be oriented weakly. 13.83/4.65 Used ordering: Polynomial interpretation [POLO]: 13.83/4.65 13.83/4.65 POL(0) = 0 13.83/4.65 POL(ACTIVE(x_1)) = x_1 13.83/4.65 POL(MARK(x_1)) = x_1 13.83/4.65 POL(active(x_1)) = x_1 13.83/4.65 POL(cons(x_1, x_2)) = x_1 13.83/4.65 POL(filter(x_1, x_2, x_3)) = x_1 + x_2 + x_3 13.83/4.65 POL(mark(x_1)) = x_1 13.83/4.65 POL(nats(x_1)) = x_1 13.83/4.65 POL(s(x_1)) = x_1 13.83/4.65 POL(sieve(x_1)) = x_1 13.83/4.65 POL(zprimes) = 1 13.83/4.65 13.83/4.65 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 13.83/4.65 13.83/4.65 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.65 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.65 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.65 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.65 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.65 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.65 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.65 mark(0) -> active(0) 13.83/4.65 mark(s(X)) -> active(s(mark(X))) 13.83/4.65 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.65 mark(zprimes) -> active(zprimes) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (44) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 ACTIVE(filter(cons(X, Y), 0, M)) -> MARK(cons(0, filter(Y, M, M))) 13.83/4.65 MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.65 MARK(filter(X1, X2, X3)) -> ACTIVE(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 ACTIVE(filter(cons(X, Y), s(N), M)) -> MARK(cons(X, filter(Y, N, M))) 13.83/4.65 ACTIVE(sieve(cons(0, Y))) -> MARK(cons(0, sieve(Y))) 13.83/4.65 ACTIVE(sieve(cons(s(N), Y))) -> MARK(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 ACTIVE(nats(N)) -> MARK(cons(N, nats(s(N)))) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X1) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X2) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X3) 13.83/4.65 MARK(s(X)) -> MARK(X) 13.83/4.65 MARK(sieve(X)) -> ACTIVE(sieve(mark(X))) 13.83/4.65 MARK(sieve(X)) -> MARK(X) 13.83/4.65 MARK(nats(X)) -> ACTIVE(nats(mark(X))) 13.83/4.65 MARK(nats(X)) -> MARK(X) 13.83/4.65 MARK(zprimes) -> ACTIVE(zprimes) 13.83/4.65 13.83/4.65 The TRS R consists of the following rules: 13.83/4.65 13.83/4.65 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.65 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.65 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.65 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.65 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.65 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.65 mark(0) -> active(0) 13.83/4.65 mark(s(X)) -> active(s(mark(X))) 13.83/4.65 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.65 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.65 mark(zprimes) -> active(zprimes) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (45) DependencyGraphProof (EQUIVALENT) 13.83/4.65 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (46) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.65 MARK(filter(X1, X2, X3)) -> ACTIVE(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 ACTIVE(filter(cons(X, Y), 0, M)) -> MARK(cons(0, filter(Y, M, M))) 13.83/4.65 ACTIVE(filter(cons(X, Y), s(N), M)) -> MARK(cons(X, filter(Y, N, M))) 13.83/4.65 ACTIVE(sieve(cons(0, Y))) -> MARK(cons(0, sieve(Y))) 13.83/4.65 ACTIVE(sieve(cons(s(N), Y))) -> MARK(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 ACTIVE(nats(N)) -> MARK(cons(N, nats(s(N)))) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X1) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X2) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X3) 13.83/4.65 MARK(s(X)) -> MARK(X) 13.83/4.65 MARK(sieve(X)) -> ACTIVE(sieve(mark(X))) 13.83/4.65 MARK(sieve(X)) -> MARK(X) 13.83/4.65 MARK(nats(X)) -> ACTIVE(nats(mark(X))) 13.83/4.65 MARK(nats(X)) -> MARK(X) 13.83/4.65 13.83/4.65 The TRS R consists of the following rules: 13.83/4.65 13.83/4.65 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.65 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.65 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.65 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.65 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.65 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.65 mark(0) -> active(0) 13.83/4.65 mark(s(X)) -> active(s(mark(X))) 13.83/4.65 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.65 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.65 mark(zprimes) -> active(zprimes) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (47) QDPOrderProof (EQUIVALENT) 13.83/4.65 We use the reduction pair processor [LPAR04,JAR06]. 13.83/4.65 13.83/4.65 13.83/4.65 The following pairs can be oriented strictly and are deleted. 13.83/4.65 13.83/4.65 ACTIVE(nats(N)) -> MARK(cons(N, nats(s(N)))) 13.83/4.65 MARK(nats(X)) -> MARK(X) 13.83/4.65 The remaining pairs can at least be oriented weakly. 13.83/4.65 Used ordering: Polynomial interpretation [POLO]: 13.83/4.65 13.83/4.65 POL(0) = 0 13.83/4.65 POL(ACTIVE(x_1)) = x_1 13.83/4.65 POL(MARK(x_1)) = x_1 13.83/4.65 POL(active(x_1)) = x_1 13.83/4.65 POL(cons(x_1, x_2)) = x_1 13.83/4.65 POL(filter(x_1, x_2, x_3)) = x_1 + x_2 + x_3 13.83/4.65 POL(mark(x_1)) = x_1 13.83/4.65 POL(nats(x_1)) = 1 + x_1 13.83/4.65 POL(s(x_1)) = x_1 13.83/4.65 POL(sieve(x_1)) = x_1 13.83/4.65 POL(zprimes) = 1 13.83/4.65 13.83/4.65 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 13.83/4.65 13.83/4.65 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.65 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.65 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.65 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.65 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.65 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.65 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.65 mark(0) -> active(0) 13.83/4.65 mark(s(X)) -> active(s(mark(X))) 13.83/4.65 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.65 mark(zprimes) -> active(zprimes) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (48) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.65 MARK(filter(X1, X2, X3)) -> ACTIVE(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 ACTIVE(filter(cons(X, Y), 0, M)) -> MARK(cons(0, filter(Y, M, M))) 13.83/4.65 ACTIVE(filter(cons(X, Y), s(N), M)) -> MARK(cons(X, filter(Y, N, M))) 13.83/4.65 ACTIVE(sieve(cons(0, Y))) -> MARK(cons(0, sieve(Y))) 13.83/4.65 ACTIVE(sieve(cons(s(N), Y))) -> MARK(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X1) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X2) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X3) 13.83/4.65 MARK(s(X)) -> MARK(X) 13.83/4.65 MARK(sieve(X)) -> ACTIVE(sieve(mark(X))) 13.83/4.65 MARK(sieve(X)) -> MARK(X) 13.83/4.65 MARK(nats(X)) -> ACTIVE(nats(mark(X))) 13.83/4.65 13.83/4.65 The TRS R consists of the following rules: 13.83/4.65 13.83/4.65 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.65 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.65 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.65 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.65 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.65 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.65 mark(0) -> active(0) 13.83/4.65 mark(s(X)) -> active(s(mark(X))) 13.83/4.65 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.65 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.65 mark(zprimes) -> active(zprimes) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (49) QDPQMonotonicMRRProof (EQUIVALENT) 13.83/4.65 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. 13.83/4.65 13.83/4.65 Strictly oriented dependency pairs: 13.83/4.65 13.83/4.65 MARK(nats(X)) -> ACTIVE(nats(mark(X))) 13.83/4.65 13.83/4.65 13.83/4.65 Used ordering: Polynomial interpretation [POLO]: 13.83/4.65 13.83/4.65 POL(0) = 0 13.83/4.65 POL(ACTIVE(x_1)) = 2*x_1 13.83/4.65 POL(MARK(x_1)) = 2 13.83/4.65 POL(active(x_1)) = 0 13.83/4.65 POL(cons(x_1, x_2)) = 0 13.83/4.65 POL(filter(x_1, x_2, x_3)) = 1 13.83/4.65 POL(mark(x_1)) = 0 13.83/4.65 POL(nats(x_1)) = 0 13.83/4.65 POL(s(x_1)) = 0 13.83/4.65 POL(sieve(x_1)) = 1 13.83/4.65 POL(zprimes) = 0 13.83/4.65 13.83/4.65 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (50) 13.83/4.65 Obligation: 13.83/4.65 Q DP problem: 13.83/4.65 The TRS P consists of the following rules: 13.83/4.65 13.83/4.65 MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.65 MARK(filter(X1, X2, X3)) -> ACTIVE(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 ACTIVE(filter(cons(X, Y), 0, M)) -> MARK(cons(0, filter(Y, M, M))) 13.83/4.65 ACTIVE(filter(cons(X, Y), s(N), M)) -> MARK(cons(X, filter(Y, N, M))) 13.83/4.65 ACTIVE(sieve(cons(0, Y))) -> MARK(cons(0, sieve(Y))) 13.83/4.65 ACTIVE(sieve(cons(s(N), Y))) -> MARK(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X1) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X2) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X3) 13.83/4.65 MARK(s(X)) -> MARK(X) 13.83/4.65 MARK(sieve(X)) -> ACTIVE(sieve(mark(X))) 13.83/4.65 MARK(sieve(X)) -> MARK(X) 13.83/4.65 13.83/4.65 The TRS R consists of the following rules: 13.83/4.65 13.83/4.65 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.65 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.65 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.65 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.65 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.65 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.65 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.65 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.65 mark(0) -> active(0) 13.83/4.65 mark(s(X)) -> active(s(mark(X))) 13.83/4.65 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.65 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.65 mark(zprimes) -> active(zprimes) 13.83/4.65 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.65 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.65 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.65 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.65 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.65 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.65 s(mark(X)) -> s(X) 13.83/4.65 s(active(X)) -> s(X) 13.83/4.65 sieve(mark(X)) -> sieve(X) 13.83/4.65 sieve(active(X)) -> sieve(X) 13.83/4.65 nats(mark(X)) -> nats(X) 13.83/4.65 nats(active(X)) -> nats(X) 13.83/4.65 13.83/4.65 The set Q consists of the following terms: 13.83/4.65 13.83/4.65 active(filter(cons(x0, x1), 0, x2)) 13.83/4.65 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.65 active(sieve(cons(0, x0))) 13.83/4.65 active(sieve(cons(s(x0), x1))) 13.83/4.65 active(nats(x0)) 13.83/4.65 active(zprimes) 13.83/4.65 mark(filter(x0, x1, x2)) 13.83/4.65 mark(cons(x0, x1)) 13.83/4.65 mark(0) 13.83/4.65 mark(s(x0)) 13.83/4.65 mark(sieve(x0)) 13.83/4.65 mark(nats(x0)) 13.83/4.65 mark(zprimes) 13.83/4.65 filter(mark(x0), x1, x2) 13.83/4.65 filter(x0, mark(x1), x2) 13.83/4.65 filter(x0, x1, mark(x2)) 13.83/4.65 filter(active(x0), x1, x2) 13.83/4.65 filter(x0, active(x1), x2) 13.83/4.65 filter(x0, x1, active(x2)) 13.83/4.65 cons(mark(x0), x1) 13.83/4.65 cons(x0, mark(x1)) 13.83/4.65 cons(active(x0), x1) 13.83/4.65 cons(x0, active(x1)) 13.83/4.65 s(mark(x0)) 13.83/4.65 s(active(x0)) 13.83/4.65 sieve(mark(x0)) 13.83/4.65 sieve(active(x0)) 13.83/4.65 nats(mark(x0)) 13.83/4.65 nats(active(x0)) 13.83/4.65 13.83/4.65 We have to consider all minimal (P,Q,R)-chains. 13.83/4.65 ---------------------------------------- 13.83/4.65 13.83/4.65 (51) QDPOrderProof (EQUIVALENT) 13.83/4.65 We use the reduction pair processor [LPAR04,JAR06]. 13.83/4.65 13.83/4.65 13.83/4.65 The following pairs can be oriented strictly and are deleted. 13.83/4.65 13.83/4.65 ACTIVE(filter(cons(X, Y), 0, M)) -> MARK(cons(0, filter(Y, M, M))) 13.83/4.65 ACTIVE(filter(cons(X, Y), s(N), M)) -> MARK(cons(X, filter(Y, N, M))) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X1) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X2) 13.83/4.65 MARK(filter(X1, X2, X3)) -> MARK(X3) 13.83/4.65 The remaining pairs can at least be oriented weakly. 13.83/4.65 Used ordering: Polynomial interpretation [POLO]: 13.83/4.65 13.83/4.65 POL(0) = 0 13.83/4.65 POL(ACTIVE(x_1)) = x_1 13.83/4.65 POL(MARK(x_1)) = x_1 13.83/4.65 POL(active(x_1)) = x_1 13.83/4.65 POL(cons(x_1, x_2)) = x_1 13.83/4.66 POL(filter(x_1, x_2, x_3)) = 1 + x_1 + x_2 + x_3 13.83/4.66 POL(mark(x_1)) = x_1 13.83/4.66 POL(nats(x_1)) = x_1 13.83/4.66 POL(s(x_1)) = x_1 13.83/4.66 POL(sieve(x_1)) = x_1 13.83/4.66 POL(zprimes) = 0 13.83/4.66 13.83/4.66 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 13.83/4.66 13.83/4.66 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.66 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.66 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.66 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.66 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.66 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.66 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.66 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.66 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.66 mark(0) -> active(0) 13.83/4.66 mark(s(X)) -> active(s(mark(X))) 13.83/4.66 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.66 mark(zprimes) -> active(zprimes) 13.83/4.66 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.66 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.66 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.66 sieve(active(X)) -> sieve(X) 13.83/4.66 sieve(mark(X)) -> sieve(X) 13.83/4.66 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.66 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.66 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.66 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.66 s(active(X)) -> s(X) 13.83/4.66 s(mark(X)) -> s(X) 13.83/4.66 nats(active(X)) -> nats(X) 13.83/4.66 nats(mark(X)) -> nats(X) 13.83/4.66 13.83/4.66 13.83/4.66 ---------------------------------------- 13.83/4.66 13.83/4.66 (52) 13.83/4.66 Obligation: 13.83/4.66 Q DP problem: 13.83/4.66 The TRS P consists of the following rules: 13.83/4.66 13.83/4.66 MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.66 MARK(filter(X1, X2, X3)) -> ACTIVE(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.66 ACTIVE(sieve(cons(0, Y))) -> MARK(cons(0, sieve(Y))) 13.83/4.66 ACTIVE(sieve(cons(s(N), Y))) -> MARK(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.66 MARK(s(X)) -> MARK(X) 13.83/4.66 MARK(sieve(X)) -> ACTIVE(sieve(mark(X))) 13.83/4.66 MARK(sieve(X)) -> MARK(X) 13.83/4.66 13.83/4.66 The TRS R consists of the following rules: 13.83/4.66 13.83/4.66 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.66 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.66 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.66 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.66 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.66 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.66 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.66 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.66 mark(0) -> active(0) 13.83/4.66 mark(s(X)) -> active(s(mark(X))) 13.83/4.66 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.66 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.66 mark(zprimes) -> active(zprimes) 13.83/4.66 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.66 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.66 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.66 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.66 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.66 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.66 s(mark(X)) -> s(X) 13.83/4.66 s(active(X)) -> s(X) 13.83/4.66 sieve(mark(X)) -> sieve(X) 13.83/4.66 sieve(active(X)) -> sieve(X) 13.83/4.66 nats(mark(X)) -> nats(X) 13.83/4.66 nats(active(X)) -> nats(X) 13.83/4.66 13.83/4.66 The set Q consists of the following terms: 13.83/4.66 13.83/4.66 active(filter(cons(x0, x1), 0, x2)) 13.83/4.66 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.66 active(sieve(cons(0, x0))) 13.83/4.66 active(sieve(cons(s(x0), x1))) 13.83/4.66 active(nats(x0)) 13.83/4.66 active(zprimes) 13.83/4.66 mark(filter(x0, x1, x2)) 13.83/4.66 mark(cons(x0, x1)) 13.83/4.66 mark(0) 13.83/4.66 mark(s(x0)) 13.83/4.66 mark(sieve(x0)) 13.83/4.66 mark(nats(x0)) 13.83/4.66 mark(zprimes) 13.83/4.66 filter(mark(x0), x1, x2) 13.83/4.66 filter(x0, mark(x1), x2) 13.83/4.66 filter(x0, x1, mark(x2)) 13.83/4.66 filter(active(x0), x1, x2) 13.83/4.66 filter(x0, active(x1), x2) 13.83/4.66 filter(x0, x1, active(x2)) 13.83/4.66 cons(mark(x0), x1) 13.83/4.66 cons(x0, mark(x1)) 13.83/4.66 cons(active(x0), x1) 13.83/4.66 cons(x0, active(x1)) 13.83/4.66 s(mark(x0)) 13.83/4.66 s(active(x0)) 13.83/4.66 sieve(mark(x0)) 13.83/4.66 sieve(active(x0)) 13.83/4.66 nats(mark(x0)) 13.83/4.66 nats(active(x0)) 13.83/4.66 13.83/4.66 We have to consider all minimal (P,Q,R)-chains. 13.83/4.66 ---------------------------------------- 13.83/4.66 13.83/4.66 (53) QDPOrderProof (EQUIVALENT) 13.83/4.66 We use the reduction pair processor [LPAR04,JAR06]. 13.83/4.66 13.83/4.66 13.83/4.66 The following pairs can be oriented strictly and are deleted. 13.83/4.66 13.83/4.66 MARK(s(X)) -> MARK(X) 13.83/4.66 The remaining pairs can at least be oriented weakly. 13.83/4.66 Used ordering: Combined order from the following AFS and order. 13.83/4.66 MARK(x1) = x1 13.83/4.66 13.83/4.66 cons(x1, x2) = x1 13.83/4.66 13.83/4.66 filter(x1, x2, x3) = x1 13.83/4.66 13.83/4.66 ACTIVE(x1) = x1 13.83/4.66 13.83/4.66 mark(x1) = x1 13.83/4.66 13.83/4.66 sieve(x1) = x1 13.83/4.66 13.83/4.66 0 = 0 13.83/4.66 13.83/4.66 s(x1) = s(x1) 13.83/4.66 13.83/4.66 active(x1) = x1 13.83/4.66 13.83/4.66 nats(x1) = x1 13.83/4.66 13.83/4.66 zprimes = zprimes 13.83/4.66 13.83/4.66 13.83/4.66 Knuth-Bendix order [KBO] with precedence:trivial 13.83/4.66 13.83/4.66 and weight map: 13.83/4.66 13.83/4.66 s_1=1 13.83/4.66 0=1 13.83/4.66 zprimes=4 13.83/4.66 13.83/4.66 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 13.83/4.66 13.83/4.66 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.66 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.66 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.66 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.66 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.66 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.66 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.66 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.66 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.66 mark(0) -> active(0) 13.83/4.66 mark(s(X)) -> active(s(mark(X))) 13.83/4.66 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.66 mark(zprimes) -> active(zprimes) 13.83/4.66 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.66 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.66 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.66 sieve(active(X)) -> sieve(X) 13.83/4.66 sieve(mark(X)) -> sieve(X) 13.83/4.66 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.66 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.66 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.66 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.66 s(active(X)) -> s(X) 13.83/4.66 s(mark(X)) -> s(X) 13.83/4.66 nats(active(X)) -> nats(X) 13.83/4.66 nats(mark(X)) -> nats(X) 13.83/4.66 13.83/4.66 13.83/4.66 ---------------------------------------- 13.83/4.66 13.83/4.66 (54) 13.83/4.66 Obligation: 13.83/4.66 Q DP problem: 13.83/4.66 The TRS P consists of the following rules: 13.83/4.66 13.83/4.66 MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.66 MARK(filter(X1, X2, X3)) -> ACTIVE(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.66 ACTIVE(sieve(cons(0, Y))) -> MARK(cons(0, sieve(Y))) 13.83/4.66 ACTIVE(sieve(cons(s(N), Y))) -> MARK(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.66 MARK(sieve(X)) -> ACTIVE(sieve(mark(X))) 13.83/4.66 MARK(sieve(X)) -> MARK(X) 13.83/4.66 13.83/4.66 The TRS R consists of the following rules: 13.83/4.66 13.83/4.66 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.66 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.66 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.66 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.66 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.66 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.66 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.66 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.66 mark(0) -> active(0) 13.83/4.66 mark(s(X)) -> active(s(mark(X))) 13.83/4.66 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.66 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.66 mark(zprimes) -> active(zprimes) 13.83/4.66 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.66 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.66 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.66 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.66 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.66 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.66 s(mark(X)) -> s(X) 13.83/4.66 s(active(X)) -> s(X) 13.83/4.66 sieve(mark(X)) -> sieve(X) 13.83/4.66 sieve(active(X)) -> sieve(X) 13.83/4.66 nats(mark(X)) -> nats(X) 13.83/4.66 nats(active(X)) -> nats(X) 13.83/4.66 13.83/4.66 The set Q consists of the following terms: 13.83/4.66 13.83/4.66 active(filter(cons(x0, x1), 0, x2)) 13.83/4.66 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.66 active(sieve(cons(0, x0))) 13.83/4.66 active(sieve(cons(s(x0), x1))) 13.83/4.66 active(nats(x0)) 13.83/4.66 active(zprimes) 13.83/4.66 mark(filter(x0, x1, x2)) 13.83/4.66 mark(cons(x0, x1)) 13.83/4.66 mark(0) 13.83/4.66 mark(s(x0)) 13.83/4.66 mark(sieve(x0)) 13.83/4.66 mark(nats(x0)) 13.83/4.66 mark(zprimes) 13.83/4.66 filter(mark(x0), x1, x2) 13.83/4.66 filter(x0, mark(x1), x2) 13.83/4.66 filter(x0, x1, mark(x2)) 13.83/4.66 filter(active(x0), x1, x2) 13.83/4.66 filter(x0, active(x1), x2) 13.83/4.66 filter(x0, x1, active(x2)) 13.83/4.66 cons(mark(x0), x1) 13.83/4.66 cons(x0, mark(x1)) 13.83/4.66 cons(active(x0), x1) 13.83/4.66 cons(x0, active(x1)) 13.83/4.66 s(mark(x0)) 13.83/4.66 s(active(x0)) 13.83/4.66 sieve(mark(x0)) 13.83/4.66 sieve(active(x0)) 13.83/4.66 nats(mark(x0)) 13.83/4.66 nats(active(x0)) 13.83/4.66 13.83/4.66 We have to consider all minimal (P,Q,R)-chains. 13.83/4.66 ---------------------------------------- 13.83/4.66 13.83/4.66 (55) QDPQMonotonicMRRProof (EQUIVALENT) 13.83/4.66 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. 13.83/4.66 13.83/4.66 Strictly oriented dependency pairs: 13.83/4.66 13.83/4.66 MARK(filter(X1, X2, X3)) -> ACTIVE(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.66 13.83/4.66 13.83/4.66 Used ordering: Polynomial interpretation [POLO]: 13.83/4.66 13.83/4.66 POL(0) = 0 13.83/4.66 POL(ACTIVE(x_1)) = x_1 13.83/4.66 POL(MARK(x_1)) = 1 13.83/4.66 POL(active(x_1)) = 0 13.83/4.66 POL(cons(x_1, x_2)) = 0 13.83/4.66 POL(filter(x_1, x_2, x_3)) = 0 13.83/4.66 POL(mark(x_1)) = 0 13.83/4.66 POL(nats(x_1)) = 0 13.83/4.66 POL(s(x_1)) = 0 13.83/4.66 POL(sieve(x_1)) = 1 13.83/4.66 POL(zprimes) = 0 13.83/4.66 13.83/4.66 13.83/4.66 ---------------------------------------- 13.83/4.66 13.83/4.66 (56) 13.83/4.66 Obligation: 13.83/4.66 Q DP problem: 13.83/4.66 The TRS P consists of the following rules: 13.83/4.66 13.83/4.66 MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.66 ACTIVE(sieve(cons(0, Y))) -> MARK(cons(0, sieve(Y))) 13.83/4.66 ACTIVE(sieve(cons(s(N), Y))) -> MARK(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.66 MARK(sieve(X)) -> ACTIVE(sieve(mark(X))) 13.83/4.66 MARK(sieve(X)) -> MARK(X) 13.83/4.66 13.83/4.66 The TRS R consists of the following rules: 13.83/4.66 13.83/4.66 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.66 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.66 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.66 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.66 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.66 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.66 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.66 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.66 mark(0) -> active(0) 13.83/4.66 mark(s(X)) -> active(s(mark(X))) 13.83/4.66 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.66 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.66 mark(zprimes) -> active(zprimes) 13.83/4.66 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.66 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.66 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.66 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.66 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.66 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.66 s(mark(X)) -> s(X) 13.83/4.66 s(active(X)) -> s(X) 13.83/4.66 sieve(mark(X)) -> sieve(X) 13.83/4.66 sieve(active(X)) -> sieve(X) 13.83/4.66 nats(mark(X)) -> nats(X) 13.83/4.66 nats(active(X)) -> nats(X) 13.83/4.66 13.83/4.66 The set Q consists of the following terms: 13.83/4.66 13.83/4.66 active(filter(cons(x0, x1), 0, x2)) 13.83/4.66 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.66 active(sieve(cons(0, x0))) 13.83/4.66 active(sieve(cons(s(x0), x1))) 13.83/4.66 active(nats(x0)) 13.83/4.66 active(zprimes) 13.83/4.66 mark(filter(x0, x1, x2)) 13.83/4.66 mark(cons(x0, x1)) 13.83/4.66 mark(0) 13.83/4.66 mark(s(x0)) 13.83/4.66 mark(sieve(x0)) 13.83/4.66 mark(nats(x0)) 13.83/4.66 mark(zprimes) 13.83/4.66 filter(mark(x0), x1, x2) 13.83/4.66 filter(x0, mark(x1), x2) 13.83/4.66 filter(x0, x1, mark(x2)) 13.83/4.66 filter(active(x0), x1, x2) 13.83/4.66 filter(x0, active(x1), x2) 13.83/4.66 filter(x0, x1, active(x2)) 13.83/4.66 cons(mark(x0), x1) 13.83/4.66 cons(x0, mark(x1)) 13.83/4.66 cons(active(x0), x1) 13.83/4.66 cons(x0, active(x1)) 13.83/4.66 s(mark(x0)) 13.83/4.66 s(active(x0)) 13.83/4.66 sieve(mark(x0)) 13.83/4.66 sieve(active(x0)) 13.83/4.66 nats(mark(x0)) 13.83/4.66 nats(active(x0)) 13.83/4.66 13.83/4.66 We have to consider all minimal (P,Q,R)-chains. 13.83/4.66 ---------------------------------------- 13.83/4.66 13.83/4.66 (57) QDPOrderProof (EQUIVALENT) 13.83/4.66 We use the reduction pair processor [LPAR04,JAR06]. 13.83/4.66 13.83/4.66 13.83/4.66 The following pairs can be oriented strictly and are deleted. 13.83/4.66 13.83/4.66 ACTIVE(sieve(cons(0, Y))) -> MARK(cons(0, sieve(Y))) 13.83/4.66 ACTIVE(sieve(cons(s(N), Y))) -> MARK(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.66 MARK(sieve(X)) -> ACTIVE(sieve(mark(X))) 13.83/4.66 MARK(sieve(X)) -> MARK(X) 13.83/4.66 The remaining pairs can at least be oriented weakly. 13.83/4.66 Used ordering: Combined order from the following AFS and order. 13.83/4.66 MARK(x1) = x1 13.83/4.66 13.83/4.66 cons(x1, x2) = x1 13.83/4.66 13.83/4.66 ACTIVE(x1) = ACTIVE 13.83/4.66 13.83/4.66 0 = 0 13.83/4.66 13.83/4.66 s(x1) = s 13.83/4.66 13.83/4.66 sieve(x1) = sieve(x1) 13.83/4.66 13.83/4.66 mark(x1) = x1 13.83/4.66 13.83/4.66 active(x1) = x1 13.83/4.66 13.83/4.66 13.83/4.66 Knuth-Bendix order [KBO] with precedence:trivial 13.83/4.66 13.83/4.66 and weight map: 13.83/4.66 13.83/4.66 s=1 13.83/4.66 0=2 13.83/4.66 ACTIVE=3 13.83/4.66 sieve_1=3 13.83/4.66 13.83/4.66 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 13.83/4.66 none 13.83/4.66 13.83/4.66 13.83/4.66 ---------------------------------------- 13.83/4.66 13.83/4.66 (58) 13.83/4.66 Obligation: 13.83/4.66 Q DP problem: 13.83/4.66 The TRS P consists of the following rules: 13.83/4.66 13.83/4.66 MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.66 13.83/4.66 The TRS R consists of the following rules: 13.83/4.66 13.83/4.66 active(filter(cons(X, Y), 0, M)) -> mark(cons(0, filter(Y, M, M))) 13.83/4.66 active(filter(cons(X, Y), s(N), M)) -> mark(cons(X, filter(Y, N, M))) 13.83/4.66 active(sieve(cons(0, Y))) -> mark(cons(0, sieve(Y))) 13.83/4.66 active(sieve(cons(s(N), Y))) -> mark(cons(s(N), sieve(filter(Y, N, N)))) 13.83/4.66 active(nats(N)) -> mark(cons(N, nats(s(N)))) 13.83/4.66 active(zprimes) -> mark(sieve(nats(s(s(0))))) 13.83/4.66 mark(filter(X1, X2, X3)) -> active(filter(mark(X1), mark(X2), mark(X3))) 13.83/4.66 mark(cons(X1, X2)) -> active(cons(mark(X1), X2)) 13.83/4.66 mark(0) -> active(0) 13.83/4.66 mark(s(X)) -> active(s(mark(X))) 13.83/4.66 mark(sieve(X)) -> active(sieve(mark(X))) 13.83/4.66 mark(nats(X)) -> active(nats(mark(X))) 13.83/4.66 mark(zprimes) -> active(zprimes) 13.83/4.66 filter(mark(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, mark(X2), X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, X2, mark(X3)) -> filter(X1, X2, X3) 13.83/4.66 filter(active(X1), X2, X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, active(X2), X3) -> filter(X1, X2, X3) 13.83/4.66 filter(X1, X2, active(X3)) -> filter(X1, X2, X3) 13.83/4.66 cons(mark(X1), X2) -> cons(X1, X2) 13.83/4.66 cons(X1, mark(X2)) -> cons(X1, X2) 13.83/4.66 cons(active(X1), X2) -> cons(X1, X2) 13.83/4.66 cons(X1, active(X2)) -> cons(X1, X2) 13.83/4.66 s(mark(X)) -> s(X) 13.83/4.66 s(active(X)) -> s(X) 13.83/4.66 sieve(mark(X)) -> sieve(X) 13.83/4.66 sieve(active(X)) -> sieve(X) 13.83/4.66 nats(mark(X)) -> nats(X) 13.83/4.66 nats(active(X)) -> nats(X) 13.83/4.66 13.83/4.66 The set Q consists of the following terms: 13.83/4.66 13.83/4.66 active(filter(cons(x0, x1), 0, x2)) 13.83/4.66 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.66 active(sieve(cons(0, x0))) 13.83/4.66 active(sieve(cons(s(x0), x1))) 13.83/4.66 active(nats(x0)) 13.83/4.66 active(zprimes) 13.83/4.66 mark(filter(x0, x1, x2)) 13.83/4.66 mark(cons(x0, x1)) 13.83/4.66 mark(0) 13.83/4.66 mark(s(x0)) 13.83/4.66 mark(sieve(x0)) 13.83/4.66 mark(nats(x0)) 13.83/4.66 mark(zprimes) 13.83/4.66 filter(mark(x0), x1, x2) 13.83/4.66 filter(x0, mark(x1), x2) 13.83/4.66 filter(x0, x1, mark(x2)) 13.83/4.66 filter(active(x0), x1, x2) 13.83/4.66 filter(x0, active(x1), x2) 13.83/4.66 filter(x0, x1, active(x2)) 13.83/4.66 cons(mark(x0), x1) 13.83/4.66 cons(x0, mark(x1)) 13.83/4.66 cons(active(x0), x1) 13.83/4.66 cons(x0, active(x1)) 13.83/4.66 s(mark(x0)) 13.83/4.66 s(active(x0)) 13.83/4.66 sieve(mark(x0)) 13.83/4.66 sieve(active(x0)) 13.83/4.66 nats(mark(x0)) 13.83/4.66 nats(active(x0)) 13.83/4.66 13.83/4.66 We have to consider all minimal (P,Q,R)-chains. 13.83/4.66 ---------------------------------------- 13.83/4.66 13.83/4.66 (59) UsableRulesProof (EQUIVALENT) 13.83/4.66 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. 13.83/4.66 ---------------------------------------- 13.83/4.66 13.83/4.66 (60) 13.83/4.66 Obligation: 13.83/4.66 Q DP problem: 13.83/4.66 The TRS P consists of the following rules: 13.83/4.66 13.83/4.66 MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.66 13.83/4.66 R is empty. 13.83/4.66 The set Q consists of the following terms: 13.83/4.66 13.83/4.66 active(filter(cons(x0, x1), 0, x2)) 13.83/4.66 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.66 active(sieve(cons(0, x0))) 13.83/4.66 active(sieve(cons(s(x0), x1))) 13.83/4.66 active(nats(x0)) 13.83/4.66 active(zprimes) 13.83/4.66 mark(filter(x0, x1, x2)) 13.83/4.66 mark(cons(x0, x1)) 13.83/4.66 mark(0) 13.83/4.66 mark(s(x0)) 13.83/4.66 mark(sieve(x0)) 13.83/4.66 mark(nats(x0)) 13.83/4.66 mark(zprimes) 13.83/4.66 filter(mark(x0), x1, x2) 13.83/4.66 filter(x0, mark(x1), x2) 13.83/4.66 filter(x0, x1, mark(x2)) 13.83/4.66 filter(active(x0), x1, x2) 13.83/4.66 filter(x0, active(x1), x2) 13.83/4.66 filter(x0, x1, active(x2)) 13.83/4.66 cons(mark(x0), x1) 13.83/4.66 cons(x0, mark(x1)) 13.83/4.66 cons(active(x0), x1) 13.83/4.66 cons(x0, active(x1)) 13.83/4.66 s(mark(x0)) 13.83/4.66 s(active(x0)) 13.83/4.66 sieve(mark(x0)) 13.83/4.66 sieve(active(x0)) 13.83/4.66 nats(mark(x0)) 13.83/4.66 nats(active(x0)) 13.83/4.66 13.83/4.66 We have to consider all minimal (P,Q,R)-chains. 13.83/4.66 ---------------------------------------- 13.83/4.66 13.83/4.66 (61) QReductionProof (EQUIVALENT) 13.83/4.66 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 13.83/4.66 13.83/4.66 active(filter(cons(x0, x1), 0, x2)) 13.83/4.66 active(filter(cons(x0, x1), s(x2), x3)) 13.83/4.66 active(sieve(cons(0, x0))) 13.83/4.66 active(sieve(cons(s(x0), x1))) 13.83/4.66 active(nats(x0)) 13.83/4.66 active(zprimes) 13.83/4.66 mark(filter(x0, x1, x2)) 13.83/4.66 mark(cons(x0, x1)) 13.83/4.66 mark(0) 13.83/4.66 mark(s(x0)) 13.83/4.66 mark(sieve(x0)) 13.83/4.66 mark(nats(x0)) 13.83/4.66 mark(zprimes) 13.83/4.66 filter(mark(x0), x1, x2) 13.83/4.66 filter(x0, mark(x1), x2) 13.83/4.66 filter(x0, x1, mark(x2)) 13.83/4.66 filter(active(x0), x1, x2) 13.83/4.66 filter(x0, active(x1), x2) 13.83/4.66 filter(x0, x1, active(x2)) 13.83/4.66 s(mark(x0)) 13.83/4.66 s(active(x0)) 13.83/4.66 sieve(mark(x0)) 13.83/4.66 sieve(active(x0)) 13.83/4.66 nats(mark(x0)) 13.83/4.66 nats(active(x0)) 13.83/4.66 13.83/4.66 13.83/4.66 ---------------------------------------- 13.83/4.66 13.83/4.66 (62) 13.83/4.66 Obligation: 13.83/4.66 Q DP problem: 13.83/4.66 The TRS P consists of the following rules: 13.83/4.66 13.83/4.66 MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.66 13.83/4.66 R is empty. 13.83/4.66 The set Q consists of the following terms: 13.83/4.66 13.83/4.66 cons(mark(x0), x1) 13.83/4.66 cons(x0, mark(x1)) 13.83/4.66 cons(active(x0), x1) 13.83/4.66 cons(x0, active(x1)) 13.83/4.66 13.83/4.66 We have to consider all minimal (P,Q,R)-chains. 13.83/4.66 ---------------------------------------- 13.83/4.66 13.83/4.66 (63) QDPSizeChangeProof (EQUIVALENT) 13.83/4.66 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. 13.83/4.66 13.83/4.66 From the DPs we obtained the following set of size-change graphs: 13.83/4.66 *MARK(cons(X1, X2)) -> MARK(X1) 13.83/4.66 The graph contains the following edges 1 > 1 13.83/4.66 13.83/4.66 13.83/4.66 ---------------------------------------- 13.83/4.66 13.83/4.66 (64) 13.83/4.66 YES 13.83/4.70 EOF