/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 134 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) RisEmptyProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) fst(pair(XS, YS)) -> XS snd(pair(XS, YS)) -> YS splitAt(0, XS) -> pair(nil, XS) splitAt(s(N), cons(X, XS)) -> u(splitAt(N, activate(XS)), N, X, activate(XS)) u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS) head(cons(N, XS)) -> N tail(cons(N, XS)) -> activate(XS) sel(N, XS) -> head(afterNth(N, XS)) take(N, XS) -> fst(splitAt(N, XS)) afterNth(N, XS) -> snd(splitAt(N, XS)) natsFrom(X) -> n__natsFrom(X) s(X) -> n__s(X) activate(n__natsFrom(X)) -> natsFrom(activate(X)) activate(n__s(X)) -> s(activate(X)) activate(X) -> X Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: natsFrom/1(YES) cons/2(YES,YES) n__natsFrom/1(YES) n__s/1(YES) fst/1)YES( pair/2(YES,YES) snd/1)YES( splitAt/2(YES,YES) 0/0) nil/0) s/1(YES) u/4(YES,YES,YES,YES) activate/1(YES) head/1)YES( tail/1(YES) sel/2(YES,YES) afterNth/2(YES,YES) take/2(YES,YES) Quasi precedence: [splitAt_2, sel_2, afterNth_2, take_2] > nil > [cons_2, n__s_1] [splitAt_2, sel_2, afterNth_2, take_2] > [u_4, activate_1, tail_1] > natsFrom_1 > n__natsFrom_1 > [cons_2, n__s_1] [splitAt_2, sel_2, afterNth_2, take_2] > [u_4, activate_1, tail_1] > pair_2 > [cons_2, n__s_1] [splitAt_2, sel_2, afterNth_2, take_2] > [u_4, activate_1, tail_1] > s_1 > [cons_2, n__s_1] 0 > pair_2 > [cons_2, n__s_1] Status: natsFrom_1: multiset status cons_2: multiset status n__natsFrom_1: multiset status n__s_1: multiset status pair_2: multiset status splitAt_2: [1,2] 0: multiset status nil: multiset status s_1: multiset status u_4: multiset status activate_1: multiset status tail_1: multiset status sel_2: [1,2] afterNth_2: [1,2] take_2: [1,2] With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: natsFrom(N) -> cons(N, n__natsFrom(n__s(N))) fst(pair(XS, YS)) -> XS snd(pair(XS, YS)) -> YS splitAt(0, XS) -> pair(nil, XS) splitAt(s(N), cons(X, XS)) -> u(splitAt(N, activate(XS)), N, X, activate(XS)) u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS) head(cons(N, XS)) -> N tail(cons(N, XS)) -> activate(XS) natsFrom(X) -> n__natsFrom(X) s(X) -> n__s(X) activate(n__natsFrom(X)) -> natsFrom(activate(X)) activate(n__s(X)) -> s(activate(X)) activate(X) -> X ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: sel(N, XS) -> head(afterNth(N, XS)) take(N, XS) -> fst(splitAt(N, XS)) afterNth(N, XS) -> snd(splitAt(N, XS)) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Knuth-Bendix order [KBO] with precedence:afterNth_2 > snd_1 > splitAt_2 > take_2 > fst_1 > sel_2 > head_1 and weight map: head_1=1 fst_1=1 snd_1=1 sel_2=2 afterNth_2=1 take_2=1 splitAt_2=0 The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: sel(N, XS) -> head(afterNth(N, XS)) take(N, XS) -> fst(splitAt(N, XS)) afterNth(N, XS) -> snd(splitAt(N, XS)) ---------------------------------------- (4) Obligation: Q restricted rewrite system: R is empty. Q is empty. ---------------------------------------- (5) RisEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. ---------------------------------------- (6) YES