details

property	value
status	complete
benchmark	quick.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n014.star.cs.uiowa.edu
space	Mixed_CTRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.06661295891 seconds
cpu usage	4.974933307
max memory	2.83910144E8

stage attributes

key	value
output-size	11406
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) QDPSizeChangeProof [EQUIVALENT, 0 ms] (7) YES (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES (11) QDP (12) QDPOrderProof [EQUIVALENT, 30 ms] (13) QDP (14) DependencyGraphProof [EQUIVALENT, 0 ms] (15) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: test(x_0, y) -> True test(x_0, y) -> False append(l1_2, l2_1) -> match_0(l1_2, l2_1, l1_2) match_0(l1_2, l2_1, Nil) -> l2_1 match_0(l1_2, l2_1, Cons(x, l)) -> Cons(x, append(l, l2_1)) part(a_4, l_3) -> match_1(a_4, l_3, l_3) match_1(a_4, l_3, Nil) -> Pair(Nil, Nil) match_1(a_4, l_3, Cons(x, l')) -> match_2(x, l', a_4, l_3, part(a_4, l')) match_2(x, l', a_4, l_3, Pair(l1, l2)) -> match_3(l1, l2, x, l', a_4, l_3, test(a_4, x)) match_3(l1, l2, x, l', a_4, l_3, False) -> Pair(Cons(x, l1), l2) match_3(l1, l2, x, l', a_4, l_3, True) -> Pair(l1, Cons(x, l2)) quick(l_5) -> match_4(l_5, l_5) match_4(l_5, Nil) -> Nil match_4(l_5, Cons(a, l')) -> match_5(a, l', l_5, part(a, l')) match_5(a, l', l_5, Pair(l1, l2)) -> append(quick(l1), Cons(a, quick(l2))) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: APPEND(l1_2, l2_1) -> MATCH_0(l1_2, l2_1, l1_2) MATCH_0(l1_2, l2_1, Cons(x, l)) -> APPEND(l, l2_1) PART(a_4, l_3) -> MATCH_1(a_4, l_3, l_3) MATCH_1(a_4, l_3, Cons(x, l')) -> MATCH_2(x, l', a_4, l_3, part(a_4, l')) MATCH_1(a_4, l_3, Cons(x, l')) -> PART(a_4, l') MATCH_2(x, l', a_4, l_3, Pair(l1, l2)) -> MATCH_3(l1, l2, x, l', a_4, l_3, test(a_4, x)) MATCH_2(x, l', a_4, l_3, Pair(l1, l2)) -> TEST(a_4, x) QUICK(l_5) -> MATCH_4(l_5, l_5) MATCH_4(l_5, Cons(a, l')) -> MATCH_5(a, l', l_5, part(a, l')) MATCH_4(l_5, Cons(a, l')) -> PART(a, l') MATCH_5(a, l', l_5, Pair(l1, l2)) -> APPEND(quick(l1), Cons(a, quick(l2))) MATCH_5(a, l', l_5, Pair(l1, l2)) -> QUICK(l1) MATCH_5(a, l', l_5, Pair(l1, l2)) -> QUICK(l2) The TRS R consists of the following rules: test(x_0, y) -> True test(x_0, y) -> False append(l1_2, l2_1) -> match_0(l1_2, l2_1, l1_2) match_0(l1_2, l2_1, Nil) -> l2_1 match_0(l1_2, l2_1, Cons(x, l)) -> Cons(x, append(l, l2_1)) part(a_4, l_3) -> match_1(a_4, l_3, l_3) match_1(a_4, l_3, Nil) -> Pair(Nil, Nil) match_1(a_4, l_3, Cons(x, l')) -> match_2(x, l', a_4, l_3, part(a_4, l')) match_2(x, l', a_4, l_3, Pair(l1, l2)) -> match_3(l1, l2, x, l', a_4, l_3, test(a_4, x)) match_3(l1, l2, x, l', a_4, l_3, False) -> Pair(Cons(x, l1), l2) match_3(l1, l2, x, l', a_4, l_3, True) -> Pair(l1, Cons(x, l2)) quick(l_5) -> match_4(l_5, l_5) match_4(l_5, Nil) -> Nil match_4(l_5, Cons(a, l')) -> match_5(a, l', l_5, part(a, l')) match_5(a, l', l_5, Pair(l1, l2)) -> append(quick(l1), Cons(a, quick(l2))) Q is empty. popout

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all output return to TRS Condi 20667