details

property	value
status	complete
benchmark	214320.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n138.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.9332 seconds
cpu usage	20.1118
user time	19.138
system time	0.973796
max virtual memory	1.9882584E7
max residence set size	2693656.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 304 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) QDPSizeChangeProof [EQUIVALENT, 0 ms] (7) YES (8) QDP (9) UsableRulesProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES (13) QDP (14) QDPOrderProof [EQUIVALENT, 160 ms] (15) QDP (16) PisEmptyProof [EQUIVALENT, 0 ms] (17) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(0(x1))) -> 0(2(1(0(x1)))) 0(1(0(x1))) -> 0(0(2(1(0(x1))))) 0(1(0(x1))) -> 0(0(2(1(2(x1))))) 0(1(0(x1))) -> 0(2(1(0(2(x1))))) 0(1(0(x1))) -> 0(3(2(1(0(x1))))) 0(1(0(x1))) -> 1(0(0(0(2(x1))))) 0(1(0(x1))) -> 1(0(0(2(0(x1))))) 0(1(0(x1))) -> 1(0(4(2(0(x1))))) 0(1(0(x1))) -> 1(4(0(4(0(x1))))) 0(1(0(x1))) -> 4(0(0(2(1(x1))))) 0(1(0(x1))) -> 5(0(0(4(1(x1))))) 0(1(0(x1))) -> 5(1(0(4(0(x1))))) 0(1(0(x1))) -> 0(2(1(0(3(2(x1)))))) 0(1(0(x1))) -> 0(4(0(4(1(3(x1)))))) 0(1(0(x1))) -> 0(4(2(2(1(0(x1)))))) 0(1(0(x1))) -> 0(5(2(1(2(0(x1)))))) 0(1(0(x1))) -> 0(5(2(5(1(0(x1)))))) 0(1(0(x1))) -> 1(0(0(5(4(4(x1)))))) 0(1(0(x1))) -> 1(0(4(4(4(0(x1)))))) 0(1(0(x1))) -> 1(5(0(0(4(2(x1)))))) 0(1(0(x1))) -> 3(0(0(4(1(4(x1)))))) 0(1(0(x1))) -> 4(5(1(0(2(0(x1)))))) 0(1(0(x1))) -> 5(5(1(0(0(2(x1)))))) 0(0(1(0(x1)))) -> 1(0(0(2(0(x1))))) 0(0(1(0(x1)))) -> 0(1(5(0(0(2(x1)))))) 0(1(0(3(x1)))) -> 1(0(3(3(0(2(x1)))))) 0(1(0(3(x1)))) -> 1(0(5(3(2(0(x1)))))) 0(1(1(0(x1)))) -> 0(4(4(1(1(0(x1)))))) 0(1(1(3(x1)))) -> 3(4(5(1(1(0(x1)))))) 0(1(2(0(x1)))) -> 1(1(0(2(0(x1))))) 0(1(2(0(x1)))) -> 3(0(2(1(0(x1))))) 0(1(2(0(x1)))) -> 4(1(0(0(2(x1))))) 0(1(2(0(x1)))) -> 0(0(4(2(5(1(x1)))))) 0(1(2(0(x1)))) -> 1(1(2(0(4(0(x1)))))) 0(1(2(0(x1)))) -> 3(0(2(1(0(4(x1)))))) 0(1(3(0(x1)))) -> 1(0(3(0(2(x1))))) 0(1(4(0(x1)))) -> 0(3(4(2(1(0(x1)))))) 0(1(5(0(x1)))) -> 0(5(1(4(0(x1))))) 0(1(5(0(x1)))) -> 1(5(3(0(2(0(x1)))))) 0(3(1(0(x1)))) -> 1(2(3(0(5(0(x1)))))) 5(0(1(0(x1)))) -> 1(4(0(0(5(1(x1)))))) 5(0(1(0(x1)))) -> 2(1(0(0(4(5(x1)))))) 0(1(0(0(0(x1))))) -> 0(0(5(1(0(0(x1)))))) 0(1(2(4(0(x1))))) -> 0(0(5(4(2(1(x1)))))) 0(1(2(5(0(x1))))) -> 1(0(2(0(5(4(x1)))))) 0(1(4(0(0(x1))))) -> 0(0(0(4(1(0(x1)))))) 0(1(4(5(0(x1))))) -> 1(5(0(0(4(2(x1)))))) 0(3(0(1(0(x1))))) -> 0(3(0(0(2(1(x1)))))) 3(0(3(1(0(x1))))) -> 0(1(3(2(3(0(x1)))))) 5(0(1(2(0(x1))))) -> 0(0(5(2(1(0(x1)))))) 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: 0^1(1(0(x1))) -> 0^1(2(1(0(x1)))) 0^1(1(0(x1))) -> 0^1(0(2(1(0(x1))))) popout

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actions

all output return to SRS Standard