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	n144.star.cs.uiowa.edu
space	Mixed_CTRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.07132 seconds
cpu usage	4.86695
user time	4.67485
system time	0.192109
max virtual memory	1.8411964E7
max residence set size	291988.0

stage attributes

key	value
starexec-result	YES

output

4.68/2.02 YES 4.68/2.03 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 4.68/2.03 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.68/2.03 4.68/2.03 4.68/2.03 Termination w.r.t. Q of the given QTRS could be proven: 4.68/2.03 4.68/2.03 (0) QTRS 4.68/2.03 (1) DependencyPairsProof [EQUIVALENT, 0 ms] 4.68/2.03 (2) QDP 4.68/2.03 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 4.68/2.03 (4) AND 4.68/2.03 (5) QDP 4.68/2.03 (6) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.68/2.03 (7) YES 4.68/2.03 (8) QDP 4.68/2.03 (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.68/2.03 (10) YES 4.68/2.03 (11) QDP 4.68/2.03 (12) QDPOrderProof [EQUIVALENT, 36 ms] 4.68/2.03 (13) QDP 4.68/2.03 (14) DependencyGraphProof [EQUIVALENT, 0 ms] 4.68/2.03 (15) TRUE 4.68/2.03 4.68/2.03 4.68/2.03 ---------------------------------------- 4.68/2.03 4.68/2.03 (0) 4.68/2.03 Obligation: 4.68/2.03 Q restricted rewrite system: 4.68/2.03 The TRS R consists of the following rules: 4.68/2.03 4.68/2.03 test(x_0, y) -> True 4.68/2.03 test(x_0, y) -> False 4.68/2.03 append(l1_2, l2_1) -> match_0(l1_2, l2_1, l1_2) 4.68/2.03 match_0(l1_2, l2_1, Nil) -> l2_1 4.68/2.03 match_0(l1_2, l2_1, Cons(x, l)) -> Cons(x, append(l, l2_1)) 4.68/2.03 part(a_4, l_3) -> match_1(a_4, l_3, l_3) 4.68/2.03 match_1(a_4, l_3, Nil) -> Pair(Nil, Nil) 4.68/2.03 match_1(a_4, l_3, Cons(x, l')) -> match_2(x, l', a_4, l_3, part(a_4, l')) 4.68/2.03 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)) 4.68/2.03 match_3(l1, l2, x, l', a_4, l_3, False) -> Pair(Cons(x, l1), l2) 4.68/2.03 match_3(l1, l2, x, l', a_4, l_3, True) -> Pair(l1, Cons(x, l2)) 4.68/2.03 quick(l_5) -> match_4(l_5, l_5) 4.68/2.03 match_4(l_5, Nil) -> Nil 4.68/2.03 match_4(l_5, Cons(a, l')) -> match_5(a, l', l_5, part(a, l')) 4.68/2.03 match_5(a, l', l_5, Pair(l1, l2)) -> append(quick(l1), Cons(a, quick(l2))) 4.68/2.03 4.68/2.03 Q is empty. 4.68/2.03 4.68/2.03 ---------------------------------------- 4.68/2.03 4.68/2.03 (1) DependencyPairsProof (EQUIVALENT) 4.68/2.03 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 4.68/2.03 ---------------------------------------- 4.68/2.03 4.68/2.03 (2) 4.68/2.03 Obligation: 4.68/2.03 Q DP problem: 4.68/2.03 The TRS P consists of the following rules: 4.68/2.03 4.68/2.03 APPEND(l1_2, l2_1) -> MATCH_0(l1_2, l2_1, l1_2) 4.68/2.03 MATCH_0(l1_2, l2_1, Cons(x, l)) -> APPEND(l, l2_1) 4.68/2.03 PART(a_4, l_3) -> MATCH_1(a_4, l_3, l_3) 4.68/2.03 MATCH_1(a_4, l_3, Cons(x, l')) -> MATCH_2(x, l', a_4, l_3, part(a_4, l')) 4.68/2.03 MATCH_1(a_4, l_3, Cons(x, l')) -> PART(a_4, l') 4.68/2.03 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)) 4.68/2.03 MATCH_2(x, l', a_4, l_3, Pair(l1, l2)) -> TEST(a_4, x) 4.68/2.03 QUICK(l_5) -> MATCH_4(l_5, l_5) 4.68/2.03 MATCH_4(l_5, Cons(a, l')) -> MATCH_5(a, l', l_5, part(a, l')) 4.68/2.03 MATCH_4(l_5, Cons(a, l')) -> PART(a, l') 4.68/2.03 MATCH_5(a, l', l_5, Pair(l1, l2)) -> APPEND(quick(l1), Cons(a, quick(l2))) 4.68/2.03 MATCH_5(a, l', l_5, Pair(l1, l2)) -> QUICK(l1) 4.68/2.03 MATCH_5(a, l', l_5, Pair(l1, l2)) -> QUICK(l2) 4.68/2.03 4.68/2.03 The TRS R consists of the following rules: 4.68/2.03 4.68/2.03 test(x_0, y) -> True 4.68/2.03 test(x_0, y) -> False 4.68/2.03 append(l1_2, l2_1) -> match_0(l1_2, l2_1, l1_2) 4.68/2.03 match_0(l1_2, l2_1, Nil) -> l2_1 4.68/2.03 match_0(l1_2, l2_1, Cons(x, l)) -> Cons(x, append(l, l2_1)) 4.68/2.03 part(a_4, l_3) -> match_1(a_4, l_3, l_3) 4.68/2.03 match_1(a_4, l_3, Nil) -> Pair(Nil, Nil) 4.68/2.03 match_1(a_4, l_3, Cons(x, l')) -> match_2(x, l', a_4, l_3, part(a_4, l')) 4.68/2.03 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)) 4.68/2.03 match_3(l1, l2, x, l', a_4, l_3, False) -> Pair(Cons(x, l1), l2) 4.68/2.03 match_3(l1, l2, x, l', a_4, l_3, True) -> Pair(l1, Cons(x, l2)) 4.68/2.03 quick(l_5) -> match_4(l_5, l_5) 4.68/2.03 match_4(l_5, Nil) -> Nil 4.68/2.03 match_4(l_5, Cons(a, l')) -> match_5(a, l', l_5, part(a, l')) 4.68/2.03 match_5(a, l', l_5, Pair(l1, l2)) -> append(quick(l1), Cons(a, quick(l2))) 4.68/2.03 4.68/2.03 Q is empty. 4.68/2.03 We have to consider all minimal (P,Q,R)-chains. 4.68/2.03 ---------------------------------------- 4.68/2.03 4.68/2.03 (3) DependencyGraphProof (EQUIVALENT) 4.68/2.03 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 5 less nodes. 4.68/2.03 ---------------------------------------- popout

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all output return to TRS_Conditional 2019-03-28 22.59