details

property	value
status	complete
benchmark	213407.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n143.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	10.1322 seconds
cpu usage	36.7197
user time	35.2482
system time	1.47147
max virtual memory	6.1023924E7
max residence set size	4080664.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 228 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) DependencyGraphProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPOrderProof [EQUIVALENT, 359 ms] (11) QDP (12) PisEmptyProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 378 ms] (18) QDP (19) PisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(1(x1))) -> 1(0(2(1(x1)))) 0(1(1(x1))) -> 1(1(0(3(2(x1))))) 0(1(1(x1))) -> 1(4(0(0(2(1(x1)))))) 0(1(4(x1))) -> 1(4(0(3(x1)))) 0(1(4(x1))) -> 4(0(2(1(x1)))) 0(1(4(x1))) -> 4(0(2(1(3(x1))))) 0(4(1(x1))) -> 0(2(4(1(x1)))) 0(4(1(x1))) -> 4(0(2(1(x1)))) 0(4(1(x1))) -> 4(0(3(1(x1)))) 0(4(1(x1))) -> 1(4(4(0(2(x1))))) 0(4(1(x1))) -> 2(1(4(0(2(x1))))) 0(5(4(x1))) -> 1(4(0(0(2(5(x1)))))) 0(5(4(x1))) -> 4(0(2(5(2(5(x1)))))) 0(5(4(x1))) -> 5(0(4(0(3(3(x1)))))) 4(2(1(x1))) -> 4(0(2(1(x1)))) 4(2(1(x1))) -> 1(2(4(0(2(x1))))) 0(1(0(4(x1)))) -> 0(0(2(2(1(4(x1)))))) 0(1(1(2(x1)))) -> 5(1(0(2(1(x1))))) 0(1(1(3(x1)))) -> 1(1(5(0(3(x1))))) 0(1(4(1(x1)))) -> 1(0(2(4(1(x1))))) 0(1(4(3(x1)))) -> 1(0(2(4(3(x1))))) 0(1(4(3(x1)))) -> 4(0(2(1(3(x1))))) 0(4(1(2(x1)))) -> 2(0(3(1(4(x1))))) 0(4(2(1(x1)))) -> 0(4(3(0(2(1(x1)))))) 0(5(0(4(x1)))) -> 1(5(4(0(0(2(x1)))))) 0(5(1(3(x1)))) -> 1(1(5(0(3(x1))))) 0(5(1(3(x1)))) -> 5(3(1(0(3(x1))))) 0(5(4(1(x1)))) -> 4(1(5(4(0(2(x1)))))) 0(5(4(3(x1)))) -> 0(2(5(0(3(4(x1)))))) 0(5(4(3(x1)))) -> 1(5(3(4(0(3(x1)))))) 1(0(5(4(x1)))) -> 1(4(5(0(3(3(x1)))))) 1(0(5(4(x1)))) -> 5(5(1(0(2(4(x1)))))) 1(4(2(1(x1)))) -> 4(0(2(1(2(1(x1)))))) 4(1(2(1(x1)))) -> 4(1(0(2(1(x1))))) 4(1(2(1(x1)))) -> 3(4(0(2(1(1(x1)))))) 4(3(2(1(x1)))) -> 0(3(4(0(2(1(x1)))))) 0(0(1(2(3(x1))))) -> 1(0(3(0(0(2(x1)))))) 0(0(5(1(2(x1))))) -> 0(0(2(5(0(1(x1)))))) 0(0(5(1(3(x1))))) -> 4(5(0(0(3(1(x1)))))) 0(1(3(4(2(x1))))) -> 1(3(4(3(0(2(x1)))))) 0(4(5(3(4(x1))))) -> 0(3(2(5(4(4(x1)))))) 0(5(0(2(2(x1))))) -> 0(0(2(5(2(4(x1)))))) 0(5(1(1(3(x1))))) -> 0(3(2(1(1(5(x1)))))) 0(5(1(4(3(x1))))) -> 3(0(2(1(5(4(x1)))))) 0(5(5(1(2(x1))))) -> 1(5(3(0(2(5(x1)))))) 1(0(1(2(4(x1))))) -> 1(1(0(2(3(4(x1)))))) 1(4(2(1(2(x1))))) -> 0(2(2(1(1(4(x1)))))) 4(0(0(5(4(x1))))) -> 5(0(0(4(4(5(x1)))))) 4(2(5(4(1(x1))))) -> 4(4(1(5(3(2(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: popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to SRS Standard